A curious geometry problem: Find the $angle OBC$
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I can not find a Method for to solve this geometry problem.I don't even know how to start.In fact, I didn't want to add my nonsensical attempts.
I looked for a similar question to this question (solved), but unfortunately, I couldn't find it. That's why I need help. I think I don't have enough mathematical information to solve this problem .
geometry trigonometry contest-math euclidean-geometry triangle
asked Dec 2 '18 at 21:35
BeginnerBeginner
348110
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|
show 4 more comments
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I can not find a Method for to solve this geometry problem.I don't even know how to start.In fact, I didn't want to add my nonsensical attempts.
I looked for a similar question to this question (solved), but unfortunately, I couldn't find it. That's why I need help. I think I don't have enough mathematical information to solve this problem .
geometry trigonometry contest-math euclidean-geometry triangle
asked Dec 2 '18 at 21:35
BeginnerBeginner
348110
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Are the commas decimal points or thousands separators?
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– saulspatz
Dec 2 '18 at 21:42
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@saulspatz decimal points of course
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– Beginner
Dec 2 '18 at 21:46
2
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I don't think the answer can be expressed in a closed form. Apparently, you have to solve for $x:=angle OBC$ from $$frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)},,$$ where $beta:=angle ABC$, $alpha:=angle OAC$, $gamma:=angle OCA$, and $theta:=angle BAC=angle ACB$. Mathematica says that $$xapprox 0.033921 approx 1.94353^circ,.$$ The given parameters are so arbitrary that I don't see any reason we should expect a nice answer.
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– Batominovski
Dec 2 '18 at 21:50
2
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However, maybe the point of this question is to illustrate how to use Ceva's Theorem in the trigonometric form.
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– Batominovski
Dec 2 '18 at 21:57
1
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Let's just say that the theorem is basically a derivative of the Law of Sines. And Google is your friend.
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– Batominovski
Dec 2 '18 at 22:09
|
show 4 more comments
$begingroup$
I can not find a Method for to solve this geometry problem.I don't even know how to start.In fact, I didn't want to add my nonsensical attempts.
I looked for a similar question to this question (solved), but unfortunately, I couldn't find it. That's why I need help. I think I don't have enough mathematical information to solve this problem .
geometry trigonometry contest-math euclidean-geometry triangle
asked Dec 2 '18 at 21:35
BeginnerBeginner
348110
$endgroup$
I can not find a Method for to solve this geometry problem.I don't even know how to start.In fact, I didn't want to add my nonsensical attempts.
I looked for a similar question to this question (solved), but unfortunately, I couldn't find it. That's why I need help. I think I don't have enough mathematical information to solve this problem .
geometry trigonometry contest-math euclidean-geometry triangle
geometry trigonometry contest-math euclidean-geometry triangle
asked Dec 2 '18 at 21:35
BeginnerBeginner
348110
asked Dec 2 '18 at 21:35
BeginnerBeginner
348110
edited Dec 4 '18 at 10:18
Beginner
asked Dec 2 '18 at 21:35
BeginnerBeginner
348110
asked Dec 2 '18 at 21:35
BeginnerBeginner
348110
asked Dec 2 '18 at 21:35
BeginnerBeginner
348110
348110
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Are the commas decimal points or thousands separators?
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– saulspatz
Dec 2 '18 at 21:42
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@saulspatz decimal points of course
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– Beginner
Dec 2 '18 at 21:46
2
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I don't think the answer can be expressed in a closed form. Apparently, you have to solve for $x:=angle OBC$ from $$frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)},,$$ where $beta:=angle ABC$, $alpha:=angle OAC$, $gamma:=angle OCA$, and $theta:=angle BAC=angle ACB$. Mathematica says that $$xapprox 0.033921 approx 1.94353^circ,.$$ The given parameters are so arbitrary that I don't see any reason we should expect a nice answer.
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– Batominovski
Dec 2 '18 at 21:50
2
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However, maybe the point of this question is to illustrate how to use Ceva's Theorem in the trigonometric form.
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– Batominovski
Dec 2 '18 at 21:57
1
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Let's just say that the theorem is basically a derivative of the Law of Sines. And Google is your friend.
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– Batominovski
Dec 2 '18 at 22:09
|
show 4 more comments
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Are the commas decimal points or thousands separators?
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– saulspatz
Dec 2 '18 at 21:42
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@saulspatz decimal points of course
$endgroup$
– Beginner
Dec 2 '18 at 21:46
2
$begingroup$
I don't think the answer can be expressed in a closed form. Apparently, you have to solve for $x:=angle OBC$ from $$frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)},,$$ where $beta:=angle ABC$, $alpha:=angle OAC$, $gamma:=angle OCA$, and $theta:=angle BAC=angle ACB$. Mathematica says that $$xapprox 0.033921 approx 1.94353^circ,.$$ The given parameters are so arbitrary that I don't see any reason we should expect a nice answer.
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– Batominovski
Dec 2 '18 at 21:50
2
$begingroup$
However, maybe the point of this question is to illustrate how to use Ceva's Theorem in the trigonometric form.
$endgroup$
– Batominovski
Dec 2 '18 at 21:57
1
$begingroup$
Let's just say that the theorem is basically a derivative of the Law of Sines. And Google is your friend.
$endgroup$
– Batominovski
Dec 2 '18 at 22:09
$begingroup$
Are the commas decimal points or thousands separators?
$endgroup$
– saulspatz
Dec 2 '18 at 21:42
$begingroup$
Are the commas decimal points or thousands separators?
$endgroup$
– saulspatz
Dec 2 '18 at 21:42
$begingroup$
@saulspatz decimal points of course
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– Beginner
Dec 2 '18 at 21:46
$begingroup$
@saulspatz decimal points of course
$endgroup$
– Beginner
Dec 2 '18 at 21:46
2
2
$begingroup$
I don't think the answer can be expressed in a closed form. Apparently, you have to solve for $x:=angle OBC$ from $$frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)},,$$ where $beta:=angle ABC$, $alpha:=angle OAC$, $gamma:=angle OCA$, and $theta:=angle BAC=angle ACB$. Mathematica says that $$xapprox 0.033921 approx 1.94353^circ,.$$ The given parameters are so arbitrary that I don't see any reason we should expect a nice answer.
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– Batominovski
Dec 2 '18 at 21:50
$begingroup$
I don't think the answer can be expressed in a closed form. Apparently, you have to solve for $x:=angle OBC$ from $$frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)},,$$ where $beta:=angle ABC$, $alpha:=angle OAC$, $gamma:=angle OCA$, and $theta:=angle BAC=angle ACB$. Mathematica says that $$xapprox 0.033921 approx 1.94353^circ,.$$ The given parameters are so arbitrary that I don't see any reason we should expect a nice answer.
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– Batominovski
Dec 2 '18 at 21:50
2
2
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However, maybe the point of this question is to illustrate how to use Ceva's Theorem in the trigonometric form.
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– Batominovski
Dec 2 '18 at 21:57
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However, maybe the point of this question is to illustrate how to use Ceva's Theorem in the trigonometric form.
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– Batominovski
Dec 2 '18 at 21:57
1
1
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Let's just say that the theorem is basically a derivative of the Law of Sines. And Google is your friend.
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– Batominovski
Dec 2 '18 at 22:09
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Let's just say that the theorem is basically a derivative of the Law of Sines. And Google is your friend.
$endgroup$
– Batominovski
Dec 2 '18 at 22:09
|
show 4 more comments
3 Answers
3
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Set $alpha:=angle OAC$, $beta:=angle ABC$,$gamma:=angle OCA$ and $theta:=angle BAC=angle ACB$.
Moreover set $a:=tan(alpha)$ and $c:=tan(gamma)$.
Now draw a perpendicular line to $overline{AC}$ from $B$ and name the meeting point $M$ ($M$ is the mid-point of $overline{AC}$). Next draw a perpendicular line to $overline{AC}$ from $O$ and name the meeting point $N$.
Set $m:=|overline{AM}|$, $h:=|overline{BM}|=d(B,overline{AC})$, $h_1:=|overline{ON}|=d(O,overline{AC})$,
$b_1:=|overline{AN}|$ and $b_2:=|overline{CN}|$.
Next we will write all distances as functions of $m$.
We have
$$
frac{h_1}{b_1}=tan(alpha),quad frac{h_1}{b_2}=tan(gamma)quadtext{and}quad b_1+b_2=2m.
$$
From this it follows (after some computations) that
$$
b_1=mfrac{2tan(gamma)}{tan(alpha)+tan(gamma)},quad
b_2=mfrac{2tan(alpha)}{tan(alpha)+tan(gamma)}quad text{and}quad
h_1=mfrac{2tan(alpha)tan(gamma)}{tan(alpha)+tan(gamma)}.
$$
We also have
$$
h=m tan(theta)quad text{and} quad
|overline{MN}|=m-b_2=mfrac{tan(gamma)-tan(alpha)}{tan(alpha)+tan(gamma)}
$$
We have two unknowns, $x_1=angle MBO$ and $x=angle OBC$, with $x=fracbeta 2-x_1$.
Now consider the right triangle with $BO$ as hypotenuse, a vertical and a horizontal side. Then for the angle $x_1=angle MBO$ we have
$$
tan(x_1)=frac{|overline{MN}|}{h-h_1}=frac{tan(gamma)-tan(alpha)}{(tan(alpha)+tan(gamma))tan(theta)-2tan(alpha)tan(gamma)}.
$$
So we arrive finally at
$$
x=fracbeta 2-x_1=fracbeta 2-arctanleft(frac{tan(gamma)-tan(alpha)}{(tan(alpha)+tan(gamma))tan(theta)-2tan(alpha)tan(gamma)}right).
$$
For the given values of $alpha$, $beta$ and $gamma$ this gives
$$
x=0.0397874,
$$
or in degrees, $x=2.27965°$.
This value is not the same as the value found by @Batominovski, but it satisfies the equality
$$
frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}, .
$$
answered Dec 5 '18 at 3:13
sansan
12.6k11332
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I found a value different $x≈4,3451027°...$
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– Beginner
Dec 5 '18 at 6:27
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For $x=4,3451027°$ we have $$frac{sin(x)}{sin(beta-x)}=0.249481ne 0.147788=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}.$$
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– san
Dec 8 '18 at 2:12
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But also for the value given by @Batominovski $x=1.94353°$ we have $$frac{sin(x)}{sin(beta-x)}=0.12871 ne 0.147788=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}.$$
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– san
Dec 8 '18 at 2:16
add a comment |
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I agree with @Batominovski that this problem, with its haphazard angle measures, is probably intended as an exercise in the trigonometric form of Ceva's Theorem:
$$frac{sinangle OAC}{sin angle OAB} cdot frac{sinangle OBA}{sinangle OBC}cdotfrac{sinangle OCB}{sinangle OCA}=1 tag{$star$}$$
We're given two of these angles ($angle OAC$ and $angle OCA$) explicitly. We're also given the isosceles triangle's vertex angle ($angle ABC$), from which we may deduce base angles $angle BAC = angle BCA$; subtracting appropriately, we may consider $angle OAB$ and $angle OCB$ known. Thus, $(star)$ effectively states
$$sin angle OBA = k sinangle OBC qquad(text{say, } sintheta = ksinphi) tag{1}$$
for a known value of $k$. But we also have $angle OBA + angle OBC = angle ABC$, another known value, so that
$$sin(angle OBA + angle OBC) = sinangle ABC qquad (sin(theta+phi)=sinpsi)tag{2}$$
"All we need to do" is solve $(1)$ and $(2)$ for $sinphi$ in terms of $k$ and $psi$. Here's a pretty slick way: simply notice that
$$
sin^2psi = sin^2theta + sin^2phi + 2 sinthetasinphicospsi tag{3}
$$
(see image below) so that, replacing $sintheta$ with $ksinphi$ and noting that all sines are positive, we readily find
$$sinphi = frac{sinpsi}{sqrt{k^2 + 1 + 2 kcospsi}}tag{4}$$
Equation $(4)$ solves the crux of the problem. Substituting-in the specific angle values is just tedium. The reader can follow the discussion under $(star)$, and/or those shown in other answers, to perform the appropriate calculations, using
$$psi := angle ABC qquad k := frac{sinangle OAB}{sinangle OAC}cdotfrac{sinangle OCA}{sinangle OCB}$$
Here's a trigonograph to demonstrate $(3)$, which amounts to applying the Law of Cosines to the (shaded) $sintheta$-$sinphi$-$sinpsi$ triangle.
answered Dec 10 '18 at 21:05
BlueBlue
48k870153
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(+1) Can you explain me where is the error in my solution math.stackexchange.com/q/3031398/548054
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– Beginner
Dec 11 '18 at 20:09
add a comment |
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Let's denote $AC = a,.,$ Since the triangle $ABC$ is isosceles
$angle BAC=angle ACB=0.5cdot(180-angle ABC) approx6.6670^circ.qquadqquadqquadqquadqquad$
$angle OCA approx6.0882^circ.qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle OACapprox4.0548^circ.qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle BCO=angle ACB-angle OCAapprox 0.5788^circ.qquadqquadqquadqquadqquadqquadqquadqquad$
$angle AOC=180^circ-angle OAC-angle OCAapprox 169.8570^circ.$
We keep an extra digit for maximum precision.
$BC=dfrac{a}{2cos(ACB)}approx0.5034a;;$because triangle $ABC$ is isosceles.$;$ Applying the law of sines for the triangle $AOC$ we get $dfrac{OC}{sin (OAC)}=dfrac{AC}{sin (AOC)},.quad$ Hence $;OC=acdotdfrac{sin (OAC)}{sin (AOC)}approx0.40152a,.;$
Using the law of cosines for $triangle BOC,$:$quad BO=sqrt{OC^2+BC^2-2cdot OC cdot BCcdot cos(BCO)}approx$
$approx 0.10198a,.;$
Then we apply the law of sines for triangle $BOC:quad$
$dfrac{OC}{sin (OBC)}=dfrac{BO}{sin (BCO)},;$ thus obtaning $;sin (OBC)=dfrac{OC}{BO}sin (BCO)approx0.039773,.;,$Finally we arrive at $angle OBC=sin^{-1}0.039773approx2.279^circ.;$ ($;$last digit is not precise, and it can be said
that $angle OBCapprox2.28^circ.)$ That's it. The problem is fully solved.
$$quad$$
$qquadqquadqquadqquadqquadqquadqquadqquad$* * * * * * * * * * * *$qquadqquadqquadqquadqquadqquadqquad$
$$quad$$
Now let's retrace all the steps and condense the whole computation into a single formula:
$$sin (OBC)=frac{OC cdot sin (BCO)}{sqrt{OC^2+BC^2-2cdot OC cdot BCcdot cos (BCO)}}=$$
$$=dfrac{acdotdfrac{sin (OAC)}{sin (AOC)}sin (BCO)}{sqrt{bigg(acdot dfrac{sin (OAC)}{sin (AOC)}bigg)^2+bigg(dfrac{a}{2cos(ACB)}bigg)^2-;2acdot dfrac{sin (OAC)}{sin (AOC)}cdot dfrac{a}{2cos(ACB)}cdot cos (BCO)}}$$
$qquadqquadqquadqquadqquadqquad$ Now we denote the angles with Greek letters
$qquadqquadqquadqquadqquadqquadqquadqquadangle ACB=alphaqquad(approx 6.6670^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquadqquadangle BCO=betaqquad(approx 0.5788^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquadqquadangle AOC=gammaqquad(approx 169.8570^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquadqquadangle OAC=phiqquad(approx 4.0548^circ)qquadqquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquad$and our expression becomes neat:
$$sin (OBC)=dfrac{acdotdfrac{sin phi}{sin gamma}sin beta}{sqrt{bigg(acdot dfrac{sin phi}{sin gamma}bigg)^2+bigg(dfrac{a}{2cosalpha}bigg)^2-;2acdot dfrac{sin phi}{sin gamma}cdot dfrac{a}{2cosalpha}cdot cos beta}}$$
This expression can be simplified by cancelling out $a$ and by dividing the numerator and denominator by the fraction $dfrac{sin phi}{sin gamma}$:
$$sin (OBC)=dfrac{sin beta}{sqrt{1+bigg(dfrac{sin gamma}{2,sin phi ,cosalpha},bigg)^2-;2cdot dfrac{sin gamma}{2,sin phi, cosalpha},cos beta}}=$$
$qquadqquadqquadqquad$ and now completing the square in the denominator
$$=dfrac{sin beta}{sqrt{bigg(dfrac{sin gamma}{2,sin phi ,cosalpha},bigg)^2-;2cdot dfrac{sin gamma}{2,sin phi, cosalpha},cos beta , + , cos^2{beta} - , cos^2{beta} , + , 1}}=$$
$qquadqquadqquadqquadqquadqquad$ remembering that $1-cos^2 beta=sin^2 beta$
$$=dfrac{sin beta}{sqrt{bigg(dfrac{sin gamma}{2,sin phi ,cosalpha}-cos{beta} bigg)^2+;sin^2beta}}=$$
$;$and then dividing the numerator and denominator by $,sin beta;$ we derive the final version of the formula
$$=dfrac{1}{sqrt{bigg(dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta} bigg)^2+;1}}=$$
$qquadqquadqquadqquad$ or $$angle OBC= sin^{-1}dfrac{1}{sqrt{bigg(dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta} bigg)^2+;1}}$$
Plugging in the angle values we once again obtain the same approximate result:
$$angle OBCapproxsin^{-1}frac{1}{sqrt{632.16}}approxsin^{-1}{0.039773}approx2.279^circ,.$$
$angle ACB=alphaqquad(approx 6.6670^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle BCO=betaqquad(approx 0.5788^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle AOC=gammaqquad(approx 169.8570^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle OAC=phiqquad(approx 4.0548^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
This radical in fact can be further simplified using the formula
$$sin {(cot^{-1}{x})}=frac{1}{sqrt{x^2+1}}:$$
$qquadqquadqquadqquadqquad$ here $;x=dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta}$
$$angle OBC=sin^{-1}{dfrac{1}{sqrt{bigg(dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta} bigg)^2+;1}}}=$$
$$sin^{-1}{frac{1}{sqrt{x^2+1}}}=sin^{-1}{(sin{(cot^{-1}{x})})}=cot^{-1}{x}=cot^{-1}{bigg(frac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta}bigg)}$$
This gives the result even faster. You get the same number (can check with a calculator) $approx2.279^circ,$
I've just turned the solution in my head through Ceva's theorem and it went without the radical (square root) i.e. you get the cotangent from a trigonometric equation resulting through the use of Ceva's theorem. This answer needs to be expanded to encompass Ceva's theorem, I guess. There's the beauty in all these interconnections. That's all. I need to add the second solution through Ceva's theorem.
$$quad$$
$qquadqquadqquadqquadqquadqquadqquadqquad$* * * * * * * * * * * *$qquadqquadqquadqquadqquadqquadqquad$
$$quad$$
Alternative solution using Ceva's theorem.
Let's write Ceva's theorem in trigonometric form:
$$frac{sin(ABO)}{sin(OBC)},frac{sin(OAC)}{sin(BAO)},frac{sin(BCO)}{sin(OCA)}=1$$
$qquadqquadqquadqquadqquadqquadangle ACB=alphaqquad(approx 6.6670^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadangle BCO=betaqquad(approx 0.5788^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadangle AOC=gammaqquad(approx 169.8570^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadangle OAC=phiqquad(approx 4.0548^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquadqquad$
We use the same denotations here but there are "more" angles now. Just like was shown in the very beginning most of these angles are obvious.
$;angle OAB=alpha-phi,,;angle OCA=alpha-beta,,;$and denoting $;angle OBC;$ (the angle we have to find) as $;x;$ it follows that $;angle ABO+angle OBC=180-2alpha;$ or $;angle ABO=180-2alpha-x,.$
Now we re-write Ceva's theorem as follows
$$frac{sin(180-2alpha-x)}{sin x},frac{sin phi}{sin(alpha-phi)},frac{sin beta}{sin(alpha-beta)}=1$$
Now we simplify the first factor of this equation:
$$frac{sin(180-2alpha-x)}{sin x}=frac{sin(2alpha+x)}{sin x}=frac{sin(2alpha)cos x + cos(2alpha)sin x}{sin x}=cot x sin (2alpha)+cos (2alpha)$$
So we have
$$[cot x,sin (2alpha)+cos (2alpha)],frac{sin phi}{sin(alpha-phi)},frac{sin beta}{sin(alpha-beta)}=1$$
$$cot x,sin (2alpha)+cos (2alpha)=,frac{sin(alpha-phi),sin(alpha-beta)}{sin phi ,sin beta}$$
$$cot x =,frac{sin(alpha-phi),sin(alpha-beta)}{sin(2alpha) sin phi ,sin beta}-cot{(2alpha)}$$
$$x =,cot^{-1}{bigg(,frac{sin(alpha-phi),sin(alpha-beta)}{sin(2alpha) sin phi ,sin beta}-cot{(2alpha)}bigg)}$$
Once again by plugging the numbers we get this value:
$$xapproxcot^{-1}{25.123}=tan^{-1}{frac{1}{25.123}}approx 2.279^circ.$$
Using Ceva's theorem gives us the solution through the inverse cotangent after solving the trivial trigonometric equation. So we don't have the radical here as opposed to when we used the law of cosines.
answered Dec 10 '18 at 18:12
Ken DracoKen Draco
56929
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Regarding your PS: Ceva's Theorem is about the concurrency of lines through the vertices of a triangle, so it has something to do with this problem. In the "classical" presentation, Ceva's concurrency condition involves ratios of lengths of segments determined where the lines cut the opposite sides; those particular ratios are irrelevant here. However, the trigonometric form gives a condition involving the ratios of sines of angles formed by those lines; those ratios are not irrelevant here.
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– Blue
Dec 10 '18 at 18:21
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@Blue Math is one whole. So everything has to do with everything. You can zigzag this problem through a lot of ways. There's no doubt about that. Ceva's theorem is a beautiful one but I see no reason whatsoever trying to drag it through this specific problem.
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– Ken Draco
Dec 10 '18 at 18:27
1
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We're given information that lets us easily determine four of the six angles formed by the lines (namely, the angles at $A$ and $C$); and we have the sum of the two remaining angles (at $B$). Trigonometric Ceva then tells us the ratio of the sines of those $B$-angles, from which we can deduce the angles themselves with just a little work. There's not much (any?) dragging here; it's kinda what Ceva was meant to do. :)
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– Blue
Dec 10 '18 at 18:38
1
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@ Blue, Very good point. Thanks. Yes, indeed. I suspected that simplification was possible. doing it through Ceva reminded me about inverse trig formulas that extend into complex variables, and I left it with a radical. Gotta remedy that at least. And yes through Ceva it looks to be just as fast or probably faster.
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– Ken Draco
Dec 10 '18 at 19:53
1
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Well, Where is the error in my solution?? math.stackexchange.com/q/3031398/548054
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– Beginner
Dec 11 '18 at 20:08
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3 Answers
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$begingroup$
Set $alpha:=angle OAC$, $beta:=angle ABC$,$gamma:=angle OCA$ and $theta:=angle BAC=angle ACB$.
Moreover set $a:=tan(alpha)$ and $c:=tan(gamma)$.
Now draw a perpendicular line to $overline{AC}$ from $B$ and name the meeting point $M$ ($M$ is the mid-point of $overline{AC}$). Next draw a perpendicular line to $overline{AC}$ from $O$ and name the meeting point $N$.
Set $m:=|overline{AM}|$, $h:=|overline{BM}|=d(B,overline{AC})$, $h_1:=|overline{ON}|=d(O,overline{AC})$,
$b_1:=|overline{AN}|$ and $b_2:=|overline{CN}|$.
Next we will write all distances as functions of $m$.
We have
$$
frac{h_1}{b_1}=tan(alpha),quad frac{h_1}{b_2}=tan(gamma)quadtext{and}quad b_1+b_2=2m.
$$
From this it follows (after some computations) that
$$
b_1=mfrac{2tan(gamma)}{tan(alpha)+tan(gamma)},quad
b_2=mfrac{2tan(alpha)}{tan(alpha)+tan(gamma)}quad text{and}quad
h_1=mfrac{2tan(alpha)tan(gamma)}{tan(alpha)+tan(gamma)}.
$$
We also have
$$
h=m tan(theta)quad text{and} quad
|overline{MN}|=m-b_2=mfrac{tan(gamma)-tan(alpha)}{tan(alpha)+tan(gamma)}
$$
We have two unknowns, $x_1=angle MBO$ and $x=angle OBC$, with $x=fracbeta 2-x_1$.
Now consider the right triangle with $BO$ as hypotenuse, a vertical and a horizontal side. Then for the angle $x_1=angle MBO$ we have
$$
tan(x_1)=frac{|overline{MN}|}{h-h_1}=frac{tan(gamma)-tan(alpha)}{(tan(alpha)+tan(gamma))tan(theta)-2tan(alpha)tan(gamma)}.
$$
So we arrive finally at
$$
x=fracbeta 2-x_1=fracbeta 2-arctanleft(frac{tan(gamma)-tan(alpha)}{(tan(alpha)+tan(gamma))tan(theta)-2tan(alpha)tan(gamma)}right).
$$
For the given values of $alpha$, $beta$ and $gamma$ this gives
$$
x=0.0397874,
$$
or in degrees, $x=2.27965°$.
This value is not the same as the value found by @Batominovski, but it satisfies the equality
$$
frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}, .
$$
answered Dec 5 '18 at 3:13
sansan
12.6k11332
$endgroup$
$begingroup$
I found a value different $x≈4,3451027°...$
$endgroup$
– Beginner
Dec 5 '18 at 6:27
$begingroup$
For $x=4,3451027°$ we have $$frac{sin(x)}{sin(beta-x)}=0.249481ne 0.147788=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}.$$
$endgroup$
– san
Dec 8 '18 at 2:12
$begingroup$
But also for the value given by @Batominovski $x=1.94353°$ we have $$frac{sin(x)}{sin(beta-x)}=0.12871 ne 0.147788=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}.$$
$endgroup$
– san
Dec 8 '18 at 2:16
add a comment |
$begingroup$
Set $alpha:=angle OAC$, $beta:=angle ABC$,$gamma:=angle OCA$ and $theta:=angle BAC=angle ACB$.
Moreover set $a:=tan(alpha)$ and $c:=tan(gamma)$.
Now draw a perpendicular line to $overline{AC}$ from $B$ and name the meeting point $M$ ($M$ is the mid-point of $overline{AC}$). Next draw a perpendicular line to $overline{AC}$ from $O$ and name the meeting point $N$.
Set $m:=|overline{AM}|$, $h:=|overline{BM}|=d(B,overline{AC})$, $h_1:=|overline{ON}|=d(O,overline{AC})$,
$b_1:=|overline{AN}|$ and $b_2:=|overline{CN}|$.
Next we will write all distances as functions of $m$.
We have
$$
frac{h_1}{b_1}=tan(alpha),quad frac{h_1}{b_2}=tan(gamma)quadtext{and}quad b_1+b_2=2m.
$$
From this it follows (after some computations) that
$$
b_1=mfrac{2tan(gamma)}{tan(alpha)+tan(gamma)},quad
b_2=mfrac{2tan(alpha)}{tan(alpha)+tan(gamma)}quad text{and}quad
h_1=mfrac{2tan(alpha)tan(gamma)}{tan(alpha)+tan(gamma)}.
$$
We also have
$$
h=m tan(theta)quad text{and} quad
|overline{MN}|=m-b_2=mfrac{tan(gamma)-tan(alpha)}{tan(alpha)+tan(gamma)}
$$
We have two unknowns, $x_1=angle MBO$ and $x=angle OBC$, with $x=fracbeta 2-x_1$.
Now consider the right triangle with $BO$ as hypotenuse, a vertical and a horizontal side. Then for the angle $x_1=angle MBO$ we have
$$
tan(x_1)=frac{|overline{MN}|}{h-h_1}=frac{tan(gamma)-tan(alpha)}{(tan(alpha)+tan(gamma))tan(theta)-2tan(alpha)tan(gamma)}.
$$
So we arrive finally at
$$
x=fracbeta 2-x_1=fracbeta 2-arctanleft(frac{tan(gamma)-tan(alpha)}{(tan(alpha)+tan(gamma))tan(theta)-2tan(alpha)tan(gamma)}right).
$$
For the given values of $alpha$, $beta$ and $gamma$ this gives
$$
x=0.0397874,
$$
or in degrees, $x=2.27965°$.
This value is not the same as the value found by @Batominovski, but it satisfies the equality
$$
frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}, .
$$
answered Dec 5 '18 at 3:13
sansan
12.6k11332
$endgroup$
$begingroup$
I found a value different $x≈4,3451027°...$
$endgroup$
– Beginner
Dec 5 '18 at 6:27
$begingroup$
For $x=4,3451027°$ we have $$frac{sin(x)}{sin(beta-x)}=0.249481ne 0.147788=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}.$$
$endgroup$
– san
Dec 8 '18 at 2:12
$begingroup$
But also for the value given by @Batominovski $x=1.94353°$ we have $$frac{sin(x)}{sin(beta-x)}=0.12871 ne 0.147788=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}.$$
$endgroup$
– san
Dec 8 '18 at 2:16
add a comment |
$begingroup$
Set $alpha:=angle OAC$, $beta:=angle ABC$,$gamma:=angle OCA$ and $theta:=angle BAC=angle ACB$.
Moreover set $a:=tan(alpha)$ and $c:=tan(gamma)$.
Now draw a perpendicular line to $overline{AC}$ from $B$ and name the meeting point $M$ ($M$ is the mid-point of $overline{AC}$). Next draw a perpendicular line to $overline{AC}$ from $O$ and name the meeting point $N$.
Set $m:=|overline{AM}|$, $h:=|overline{BM}|=d(B,overline{AC})$, $h_1:=|overline{ON}|=d(O,overline{AC})$,
$b_1:=|overline{AN}|$ and $b_2:=|overline{CN}|$.
Next we will write all distances as functions of $m$.
We have
$$
frac{h_1}{b_1}=tan(alpha),quad frac{h_1}{b_2}=tan(gamma)quadtext{and}quad b_1+b_2=2m.
$$
From this it follows (after some computations) that
$$
b_1=mfrac{2tan(gamma)}{tan(alpha)+tan(gamma)},quad
b_2=mfrac{2tan(alpha)}{tan(alpha)+tan(gamma)}quad text{and}quad
h_1=mfrac{2tan(alpha)tan(gamma)}{tan(alpha)+tan(gamma)}.
$$
We also have
$$
h=m tan(theta)quad text{and} quad
|overline{MN}|=m-b_2=mfrac{tan(gamma)-tan(alpha)}{tan(alpha)+tan(gamma)}
$$
We have two unknowns, $x_1=angle MBO$ and $x=angle OBC$, with $x=fracbeta 2-x_1$.
Now consider the right triangle with $BO$ as hypotenuse, a vertical and a horizontal side. Then for the angle $x_1=angle MBO$ we have
$$
tan(x_1)=frac{|overline{MN}|}{h-h_1}=frac{tan(gamma)-tan(alpha)}{(tan(alpha)+tan(gamma))tan(theta)-2tan(alpha)tan(gamma)}.
$$
So we arrive finally at
$$
x=fracbeta 2-x_1=fracbeta 2-arctanleft(frac{tan(gamma)-tan(alpha)}{(tan(alpha)+tan(gamma))tan(theta)-2tan(alpha)tan(gamma)}right).
$$
For the given values of $alpha$, $beta$ and $gamma$ this gives
$$
x=0.0397874,
$$
or in degrees, $x=2.27965°$.
This value is not the same as the value found by @Batominovski, but it satisfies the equality
$$
frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}, .
$$
answered Dec 5 '18 at 3:13
sansan
12.6k11332
$endgroup$
Set $alpha:=angle OAC$, $beta:=angle ABC$,$gamma:=angle OCA$ and $theta:=angle BAC=angle ACB$.
Moreover set $a:=tan(alpha)$ and $c:=tan(gamma)$.
Now draw a perpendicular line to $overline{AC}$ from $B$ and name the meeting point $M$ ($M$ is the mid-point of $overline{AC}$). Next draw a perpendicular line to $overline{AC}$ from $O$ and name the meeting point $N$.
Set $m:=|overline{AM}|$, $h:=|overline{BM}|=d(B,overline{AC})$, $h_1:=|overline{ON}|=d(O,overline{AC})$,
$b_1:=|overline{AN}|$ and $b_2:=|overline{CN}|$.
Next we will write all distances as functions of $m$.
We have
$$
frac{h_1}{b_1}=tan(alpha),quad frac{h_1}{b_2}=tan(gamma)quadtext{and}quad b_1+b_2=2m.
$$
From this it follows (after some computations) that
$$
b_1=mfrac{2tan(gamma)}{tan(alpha)+tan(gamma)},quad
b_2=mfrac{2tan(alpha)}{tan(alpha)+tan(gamma)}quad text{and}quad
h_1=mfrac{2tan(alpha)tan(gamma)}{tan(alpha)+tan(gamma)}.
$$
We also have
$$
h=m tan(theta)quad text{and} quad
|overline{MN}|=m-b_2=mfrac{tan(gamma)-tan(alpha)}{tan(alpha)+tan(gamma)}
$$
We have two unknowns, $x_1=angle MBO$ and $x=angle OBC$, with $x=fracbeta 2-x_1$.
Now consider the right triangle with $BO$ as hypotenuse, a vertical and a horizontal side. Then for the angle $x_1=angle MBO$ we have
$$
tan(x_1)=frac{|overline{MN}|}{h-h_1}=frac{tan(gamma)-tan(alpha)}{(tan(alpha)+tan(gamma))tan(theta)-2tan(alpha)tan(gamma)}.
$$
So we arrive finally at
$$
x=fracbeta 2-x_1=fracbeta 2-arctanleft(frac{tan(gamma)-tan(alpha)}{(tan(alpha)+tan(gamma))tan(theta)-2tan(alpha)tan(gamma)}right).
$$
For the given values of $alpha$, $beta$ and $gamma$ this gives
$$
x=0.0397874,
$$
or in degrees, $x=2.27965°$.
This value is not the same as the value found by @Batominovski, but it satisfies the equality
$$
frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}, .
$$
answered Dec 5 '18 at 3:13
sansan
12.6k11332
edited Dec 5 '18 at 3:19
answered Dec 5 '18 at 3:13
sansan
12.6k11332
answered Dec 5 '18 at 3:13
sansan
12.6k11332
answered Dec 5 '18 at 3:13
sansan
12.6k11332
12.6k11332
$begingroup$
I found a value different $x≈4,3451027°...$
$endgroup$
– Beginner
Dec 5 '18 at 6:27
$begingroup$
For $x=4,3451027°$ we have $$frac{sin(x)}{sin(beta-x)}=0.249481ne 0.147788=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}.$$
$endgroup$
– san
Dec 8 '18 at 2:12
$begingroup$
But also for the value given by @Batominovski $x=1.94353°$ we have $$frac{sin(x)}{sin(beta-x)}=0.12871 ne 0.147788=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}.$$
$endgroup$
– san
Dec 8 '18 at 2:16
add a comment |
$begingroup$
I found a value different $x≈4,3451027°...$
$endgroup$
– Beginner
Dec 5 '18 at 6:27
$begingroup$
For $x=4,3451027°$ we have $$frac{sin(x)}{sin(beta-x)}=0.249481ne 0.147788=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}.$$
$endgroup$
– san
Dec 8 '18 at 2:12
$begingroup$
But also for the value given by @Batominovski $x=1.94353°$ we have $$frac{sin(x)}{sin(beta-x)}=0.12871 ne 0.147788=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}.$$
$endgroup$
– san
Dec 8 '18 at 2:16
$begingroup$
I found a value different $x≈4,3451027°...$
$endgroup$
– Beginner
Dec 5 '18 at 6:27
$begingroup$
I found a value different $x≈4,3451027°...$
$endgroup$
– Beginner
Dec 5 '18 at 6:27
$begingroup$
For $x=4,3451027°$ we have $$frac{sin(x)}{sin(beta-x)}=0.249481ne 0.147788=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}.$$
$endgroup$
– san
Dec 8 '18 at 2:12
$begingroup$
For $x=4,3451027°$ we have $$frac{sin(x)}{sin(beta-x)}=0.249481ne 0.147788=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}.$$
$endgroup$
– san
Dec 8 '18 at 2:12
$begingroup$
But also for the value given by @Batominovski $x=1.94353°$ we have $$frac{sin(x)}{sin(beta-x)}=0.12871 ne 0.147788=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}.$$
$endgroup$
– san
Dec 8 '18 at 2:16
$begingroup$
But also for the value given by @Batominovski $x=1.94353°$ we have $$frac{sin(x)}{sin(beta-x)}=0.12871 ne 0.147788=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}.$$
$endgroup$
– san
Dec 8 '18 at 2:16
add a comment |
$begingroup$
I agree with @Batominovski that this problem, with its haphazard angle measures, is probably intended as an exercise in the trigonometric form of Ceva's Theorem:
$$frac{sinangle OAC}{sin angle OAB} cdot frac{sinangle OBA}{sinangle OBC}cdotfrac{sinangle OCB}{sinangle OCA}=1 tag{$star$}$$
We're given two of these angles ($angle OAC$ and $angle OCA$) explicitly. We're also given the isosceles triangle's vertex angle ($angle ABC$), from which we may deduce base angles $angle BAC = angle BCA$; subtracting appropriately, we may consider $angle OAB$ and $angle OCB$ known. Thus, $(star)$ effectively states
$$sin angle OBA = k sinangle OBC qquad(text{say, } sintheta = ksinphi) tag{1}$$
for a known value of $k$. But we also have $angle OBA + angle OBC = angle ABC$, another known value, so that
$$sin(angle OBA + angle OBC) = sinangle ABC qquad (sin(theta+phi)=sinpsi)tag{2}$$
"All we need to do" is solve $(1)$ and $(2)$ for $sinphi$ in terms of $k$ and $psi$. Here's a pretty slick way: simply notice that
$$
sin^2psi = sin^2theta + sin^2phi + 2 sinthetasinphicospsi tag{3}
$$
(see image below) so that, replacing $sintheta$ with $ksinphi$ and noting that all sines are positive, we readily find
$$sinphi = frac{sinpsi}{sqrt{k^2 + 1 + 2 kcospsi}}tag{4}$$
Equation $(4)$ solves the crux of the problem. Substituting-in the specific angle values is just tedium. The reader can follow the discussion under $(star)$, and/or those shown in other answers, to perform the appropriate calculations, using
$$psi := angle ABC qquad k := frac{sinangle OAB}{sinangle OAC}cdotfrac{sinangle OCA}{sinangle OCB}$$
Here's a trigonograph to demonstrate $(3)$, which amounts to applying the Law of Cosines to the (shaded) $sintheta$-$sinphi$-$sinpsi$ triangle.
answered Dec 10 '18 at 21:05
BlueBlue
48k870153
$endgroup$
$begingroup$
(+1) Can you explain me where is the error in my solution math.stackexchange.com/q/3031398/548054
$endgroup$
– Beginner
Dec 11 '18 at 20:09
add a comment |
$begingroup$
I agree with @Batominovski that this problem, with its haphazard angle measures, is probably intended as an exercise in the trigonometric form of Ceva's Theorem:
$$frac{sinangle OAC}{sin angle OAB} cdot frac{sinangle OBA}{sinangle OBC}cdotfrac{sinangle OCB}{sinangle OCA}=1 tag{$star$}$$
We're given two of these angles ($angle OAC$ and $angle OCA$) explicitly. We're also given the isosceles triangle's vertex angle ($angle ABC$), from which we may deduce base angles $angle BAC = angle BCA$; subtracting appropriately, we may consider $angle OAB$ and $angle OCB$ known. Thus, $(star)$ effectively states
$$sin angle OBA = k sinangle OBC qquad(text{say, } sintheta = ksinphi) tag{1}$$
for a known value of $k$. But we also have $angle OBA + angle OBC = angle ABC$, another known value, so that
$$sin(angle OBA + angle OBC) = sinangle ABC qquad (sin(theta+phi)=sinpsi)tag{2}$$
"All we need to do" is solve $(1)$ and $(2)$ for $sinphi$ in terms of $k$ and $psi$. Here's a pretty slick way: simply notice that
$$
sin^2psi = sin^2theta + sin^2phi + 2 sinthetasinphicospsi tag{3}
$$
(see image below) so that, replacing $sintheta$ with $ksinphi$ and noting that all sines are positive, we readily find
$$sinphi = frac{sinpsi}{sqrt{k^2 + 1 + 2 kcospsi}}tag{4}$$
Equation $(4)$ solves the crux of the problem. Substituting-in the specific angle values is just tedium. The reader can follow the discussion under $(star)$, and/or those shown in other answers, to perform the appropriate calculations, using
$$psi := angle ABC qquad k := frac{sinangle OAB}{sinangle OAC}cdotfrac{sinangle OCA}{sinangle OCB}$$
Here's a trigonograph to demonstrate $(3)$, which amounts to applying the Law of Cosines to the (shaded) $sintheta$-$sinphi$-$sinpsi$ triangle.
answered Dec 10 '18 at 21:05
BlueBlue
48k870153
$endgroup$
$begingroup$
(+1) Can you explain me where is the error in my solution math.stackexchange.com/q/3031398/548054
$endgroup$
– Beginner
Dec 11 '18 at 20:09
add a comment |
$begingroup$
I agree with @Batominovski that this problem, with its haphazard angle measures, is probably intended as an exercise in the trigonometric form of Ceva's Theorem:
$$frac{sinangle OAC}{sin angle OAB} cdot frac{sinangle OBA}{sinangle OBC}cdotfrac{sinangle OCB}{sinangle OCA}=1 tag{$star$}$$
We're given two of these angles ($angle OAC$ and $angle OCA$) explicitly. We're also given the isosceles triangle's vertex angle ($angle ABC$), from which we may deduce base angles $angle BAC = angle BCA$; subtracting appropriately, we may consider $angle OAB$ and $angle OCB$ known. Thus, $(star)$ effectively states
$$sin angle OBA = k sinangle OBC qquad(text{say, } sintheta = ksinphi) tag{1}$$
for a known value of $k$. But we also have $angle OBA + angle OBC = angle ABC$, another known value, so that
$$sin(angle OBA + angle OBC) = sinangle ABC qquad (sin(theta+phi)=sinpsi)tag{2}$$
"All we need to do" is solve $(1)$ and $(2)$ for $sinphi$ in terms of $k$ and $psi$. Here's a pretty slick way: simply notice that
$$
sin^2psi = sin^2theta + sin^2phi + 2 sinthetasinphicospsi tag{3}
$$
(see image below) so that, replacing $sintheta$ with $ksinphi$ and noting that all sines are positive, we readily find
$$sinphi = frac{sinpsi}{sqrt{k^2 + 1 + 2 kcospsi}}tag{4}$$
Equation $(4)$ solves the crux of the problem. Substituting-in the specific angle values is just tedium. The reader can follow the discussion under $(star)$, and/or those shown in other answers, to perform the appropriate calculations, using
$$psi := angle ABC qquad k := frac{sinangle OAB}{sinangle OAC}cdotfrac{sinangle OCA}{sinangle OCB}$$
Here's a trigonograph to demonstrate $(3)$, which amounts to applying the Law of Cosines to the (shaded) $sintheta$-$sinphi$-$sinpsi$ triangle.
answered Dec 10 '18 at 21:05
BlueBlue
48k870153
$endgroup$
I agree with @Batominovski that this problem, with its haphazard angle measures, is probably intended as an exercise in the trigonometric form of Ceva's Theorem:
$$frac{sinangle OAC}{sin angle OAB} cdot frac{sinangle OBA}{sinangle OBC}cdotfrac{sinangle OCB}{sinangle OCA}=1 tag{$star$}$$
We're given two of these angles ($angle OAC$ and $angle OCA$) explicitly. We're also given the isosceles triangle's vertex angle ($angle ABC$), from which we may deduce base angles $angle BAC = angle BCA$; subtracting appropriately, we may consider $angle OAB$ and $angle OCB$ known. Thus, $(star)$ effectively states
$$sin angle OBA = k sinangle OBC qquad(text{say, } sintheta = ksinphi) tag{1}$$
for a known value of $k$. But we also have $angle OBA + angle OBC = angle ABC$, another known value, so that
$$sin(angle OBA + angle OBC) = sinangle ABC qquad (sin(theta+phi)=sinpsi)tag{2}$$
"All we need to do" is solve $(1)$ and $(2)$ for $sinphi$ in terms of $k$ and $psi$. Here's a pretty slick way: simply notice that
$$
sin^2psi = sin^2theta + sin^2phi + 2 sinthetasinphicospsi tag{3}
$$
(see image below) so that, replacing $sintheta$ with $ksinphi$ and noting that all sines are positive, we readily find
$$sinphi = frac{sinpsi}{sqrt{k^2 + 1 + 2 kcospsi}}tag{4}$$
Equation $(4)$ solves the crux of the problem. Substituting-in the specific angle values is just tedium. The reader can follow the discussion under $(star)$, and/or those shown in other answers, to perform the appropriate calculations, using
$$psi := angle ABC qquad k := frac{sinangle OAB}{sinangle OAC}cdotfrac{sinangle OCA}{sinangle OCB}$$
Here's a trigonograph to demonstrate $(3)$, which amounts to applying the Law of Cosines to the (shaded) $sintheta$-$sinphi$-$sinpsi$ triangle.
answered Dec 10 '18 at 21:05
BlueBlue
48k870153
edited Dec 11 '18 at 9:20
answered Dec 10 '18 at 21:05
BlueBlue
48k870153
answered Dec 10 '18 at 21:05
BlueBlue
48k870153
answered Dec 10 '18 at 21:05
BlueBlue
48k870153
48k870153
$begingroup$
(+1) Can you explain me where is the error in my solution math.stackexchange.com/q/3031398/548054
$endgroup$
– Beginner
Dec 11 '18 at 20:09
add a comment |
$begingroup$
(+1) Can you explain me where is the error in my solution math.stackexchange.com/q/3031398/548054
$endgroup$
– Beginner
Dec 11 '18 at 20:09
$begingroup$
(+1) Can you explain me where is the error in my solution math.stackexchange.com/q/3031398/548054
$endgroup$
– Beginner
Dec 11 '18 at 20:09
$begingroup$
(+1) Can you explain me where is the error in my solution math.stackexchange.com/q/3031398/548054
$endgroup$
– Beginner
Dec 11 '18 at 20:09
add a comment |
$begingroup$
Let's denote $AC = a,.,$ Since the triangle $ABC$ is isosceles
$angle BAC=angle ACB=0.5cdot(180-angle ABC) approx6.6670^circ.qquadqquadqquadqquadqquad$
$angle OCA approx6.0882^circ.qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle OACapprox4.0548^circ.qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle BCO=angle ACB-angle OCAapprox 0.5788^circ.qquadqquadqquadqquadqquadqquadqquadqquad$
$angle AOC=180^circ-angle OAC-angle OCAapprox 169.8570^circ.$
We keep an extra digit for maximum precision.
$BC=dfrac{a}{2cos(ACB)}approx0.5034a;;$because triangle $ABC$ is isosceles.$;$ Applying the law of sines for the triangle $AOC$ we get $dfrac{OC}{sin (OAC)}=dfrac{AC}{sin (AOC)},.quad$ Hence $;OC=acdotdfrac{sin (OAC)}{sin (AOC)}approx0.40152a,.;$
Using the law of cosines for $triangle BOC,$:$quad BO=sqrt{OC^2+BC^2-2cdot OC cdot BCcdot cos(BCO)}approx$
$approx 0.10198a,.;$
Then we apply the law of sines for triangle $BOC:quad$
$dfrac{OC}{sin (OBC)}=dfrac{BO}{sin (BCO)},;$ thus obtaning $;sin (OBC)=dfrac{OC}{BO}sin (BCO)approx0.039773,.;,$Finally we arrive at $angle OBC=sin^{-1}0.039773approx2.279^circ.;$ ($;$last digit is not precise, and it can be said
that $angle OBCapprox2.28^circ.)$ That's it. The problem is fully solved.
$$quad$$
$qquadqquadqquadqquadqquadqquadqquadqquad$* * * * * * * * * * * *$qquadqquadqquadqquadqquadqquadqquad$
$$quad$$
Now let's retrace all the steps and condense the whole computation into a single formula:
$$sin (OBC)=frac{OC cdot sin (BCO)}{sqrt{OC^2+BC^2-2cdot OC cdot BCcdot cos (BCO)}}=$$
$$=dfrac{acdotdfrac{sin (OAC)}{sin (AOC)}sin (BCO)}{sqrt{bigg(acdot dfrac{sin (OAC)}{sin (AOC)}bigg)^2+bigg(dfrac{a}{2cos(ACB)}bigg)^2-;2acdot dfrac{sin (OAC)}{sin (AOC)}cdot dfrac{a}{2cos(ACB)}cdot cos (BCO)}}$$
$qquadqquadqquadqquadqquadqquad$ Now we denote the angles with Greek letters
$qquadqquadqquadqquadqquadqquadqquadqquadangle ACB=alphaqquad(approx 6.6670^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquadqquadangle BCO=betaqquad(approx 0.5788^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquadqquadangle AOC=gammaqquad(approx 169.8570^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquadqquadangle OAC=phiqquad(approx 4.0548^circ)qquadqquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquad$and our expression becomes neat:
$$sin (OBC)=dfrac{acdotdfrac{sin phi}{sin gamma}sin beta}{sqrt{bigg(acdot dfrac{sin phi}{sin gamma}bigg)^2+bigg(dfrac{a}{2cosalpha}bigg)^2-;2acdot dfrac{sin phi}{sin gamma}cdot dfrac{a}{2cosalpha}cdot cos beta}}$$
This expression can be simplified by cancelling out $a$ and by dividing the numerator and denominator by the fraction $dfrac{sin phi}{sin gamma}$:
$$sin (OBC)=dfrac{sin beta}{sqrt{1+bigg(dfrac{sin gamma}{2,sin phi ,cosalpha},bigg)^2-;2cdot dfrac{sin gamma}{2,sin phi, cosalpha},cos beta}}=$$
$qquadqquadqquadqquad$ and now completing the square in the denominator
$$=dfrac{sin beta}{sqrt{bigg(dfrac{sin gamma}{2,sin phi ,cosalpha},bigg)^2-;2cdot dfrac{sin gamma}{2,sin phi, cosalpha},cos beta , + , cos^2{beta} - , cos^2{beta} , + , 1}}=$$
$qquadqquadqquadqquadqquadqquad$ remembering that $1-cos^2 beta=sin^2 beta$
$$=dfrac{sin beta}{sqrt{bigg(dfrac{sin gamma}{2,sin phi ,cosalpha}-cos{beta} bigg)^2+;sin^2beta}}=$$
$;$and then dividing the numerator and denominator by $,sin beta;$ we derive the final version of the formula
$$=dfrac{1}{sqrt{bigg(dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta} bigg)^2+;1}}=$$
$qquadqquadqquadqquad$ or $$angle OBC= sin^{-1}dfrac{1}{sqrt{bigg(dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta} bigg)^2+;1}}$$
Plugging in the angle values we once again obtain the same approximate result:
$$angle OBCapproxsin^{-1}frac{1}{sqrt{632.16}}approxsin^{-1}{0.039773}approx2.279^circ,.$$
$angle ACB=alphaqquad(approx 6.6670^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle BCO=betaqquad(approx 0.5788^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle AOC=gammaqquad(approx 169.8570^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle OAC=phiqquad(approx 4.0548^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
This radical in fact can be further simplified using the formula
$$sin {(cot^{-1}{x})}=frac{1}{sqrt{x^2+1}}:$$
$qquadqquadqquadqquadqquad$ here $;x=dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta}$
$$angle OBC=sin^{-1}{dfrac{1}{sqrt{bigg(dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta} bigg)^2+;1}}}=$$
$$sin^{-1}{frac{1}{sqrt{x^2+1}}}=sin^{-1}{(sin{(cot^{-1}{x})})}=cot^{-1}{x}=cot^{-1}{bigg(frac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta}bigg)}$$
This gives the result even faster. You get the same number (can check with a calculator) $approx2.279^circ,$
I've just turned the solution in my head through Ceva's theorem and it went without the radical (square root) i.e. you get the cotangent from a trigonometric equation resulting through the use of Ceva's theorem. This answer needs to be expanded to encompass Ceva's theorem, I guess. There's the beauty in all these interconnections. That's all. I need to add the second solution through Ceva's theorem.
$$quad$$
$qquadqquadqquadqquadqquadqquadqquadqquad$* * * * * * * * * * * *$qquadqquadqquadqquadqquadqquadqquad$
$$quad$$
Alternative solution using Ceva's theorem.
Let's write Ceva's theorem in trigonometric form:
$$frac{sin(ABO)}{sin(OBC)},frac{sin(OAC)}{sin(BAO)},frac{sin(BCO)}{sin(OCA)}=1$$
$qquadqquadqquadqquadqquadqquadangle ACB=alphaqquad(approx 6.6670^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadangle BCO=betaqquad(approx 0.5788^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadangle AOC=gammaqquad(approx 169.8570^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadangle OAC=phiqquad(approx 4.0548^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquadqquad$
We use the same denotations here but there are "more" angles now. Just like was shown in the very beginning most of these angles are obvious.
$;angle OAB=alpha-phi,,;angle OCA=alpha-beta,,;$and denoting $;angle OBC;$ (the angle we have to find) as $;x;$ it follows that $;angle ABO+angle OBC=180-2alpha;$ or $;angle ABO=180-2alpha-x,.$
Now we re-write Ceva's theorem as follows
$$frac{sin(180-2alpha-x)}{sin x},frac{sin phi}{sin(alpha-phi)},frac{sin beta}{sin(alpha-beta)}=1$$
Now we simplify the first factor of this equation:
$$frac{sin(180-2alpha-x)}{sin x}=frac{sin(2alpha+x)}{sin x}=frac{sin(2alpha)cos x + cos(2alpha)sin x}{sin x}=cot x sin (2alpha)+cos (2alpha)$$
So we have
$$[cot x,sin (2alpha)+cos (2alpha)],frac{sin phi}{sin(alpha-phi)},frac{sin beta}{sin(alpha-beta)}=1$$
$$cot x,sin (2alpha)+cos (2alpha)=,frac{sin(alpha-phi),sin(alpha-beta)}{sin phi ,sin beta}$$
$$cot x =,frac{sin(alpha-phi),sin(alpha-beta)}{sin(2alpha) sin phi ,sin beta}-cot{(2alpha)}$$
$$x =,cot^{-1}{bigg(,frac{sin(alpha-phi),sin(alpha-beta)}{sin(2alpha) sin phi ,sin beta}-cot{(2alpha)}bigg)}$$
Once again by plugging the numbers we get this value:
$$xapproxcot^{-1}{25.123}=tan^{-1}{frac{1}{25.123}}approx 2.279^circ.$$
Using Ceva's theorem gives us the solution through the inverse cotangent after solving the trivial trigonometric equation. So we don't have the radical here as opposed to when we used the law of cosines.
answered Dec 10 '18 at 18:12
Ken DracoKen Draco
56929
$endgroup$
$begingroup$
Regarding your PS: Ceva's Theorem is about the concurrency of lines through the vertices of a triangle, so it has something to do with this problem. In the "classical" presentation, Ceva's concurrency condition involves ratios of lengths of segments determined where the lines cut the opposite sides; those particular ratios are irrelevant here. However, the trigonometric form gives a condition involving the ratios of sines of angles formed by those lines; those ratios are not irrelevant here.
$endgroup$
– Blue
Dec 10 '18 at 18:21
$begingroup$
@Blue Math is one whole. So everything has to do with everything. You can zigzag this problem through a lot of ways. There's no doubt about that. Ceva's theorem is a beautiful one but I see no reason whatsoever trying to drag it through this specific problem.
$endgroup$
– Ken Draco
Dec 10 '18 at 18:27
1
$begingroup$
We're given information that lets us easily determine four of the six angles formed by the lines (namely, the angles at $A$ and $C$); and we have the sum of the two remaining angles (at $B$). Trigonometric Ceva then tells us the ratio of the sines of those $B$-angles, from which we can deduce the angles themselves with just a little work. There's not much (any?) dragging here; it's kinda what Ceva was meant to do. :)
$endgroup$
– Blue
Dec 10 '18 at 18:38
1
$begingroup$
@ Blue, Very good point. Thanks. Yes, indeed. I suspected that simplification was possible. doing it through Ceva reminded me about inverse trig formulas that extend into complex variables, and I left it with a radical. Gotta remedy that at least. And yes through Ceva it looks to be just as fast or probably faster.
$endgroup$
– Ken Draco
Dec 10 '18 at 19:53
1
$begingroup$
Well, Where is the error in my solution?? math.stackexchange.com/q/3031398/548054
$endgroup$
– Beginner
Dec 11 '18 at 20:08
|
show 2 more comments
$begingroup$
Let's denote $AC = a,.,$ Since the triangle $ABC$ is isosceles
$angle BAC=angle ACB=0.5cdot(180-angle ABC) approx6.6670^circ.qquadqquadqquadqquadqquad$
$angle OCA approx6.0882^circ.qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle OACapprox4.0548^circ.qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle BCO=angle ACB-angle OCAapprox 0.5788^circ.qquadqquadqquadqquadqquadqquadqquadqquad$
$angle AOC=180^circ-angle OAC-angle OCAapprox 169.8570^circ.$
We keep an extra digit for maximum precision.
$BC=dfrac{a}{2cos(ACB)}approx0.5034a;;$because triangle $ABC$ is isosceles.$;$ Applying the law of sines for the triangle $AOC$ we get $dfrac{OC}{sin (OAC)}=dfrac{AC}{sin (AOC)},.quad$ Hence $;OC=acdotdfrac{sin (OAC)}{sin (AOC)}approx0.40152a,.;$
Using the law of cosines for $triangle BOC,$:$quad BO=sqrt{OC^2+BC^2-2cdot OC cdot BCcdot cos(BCO)}approx$
$approx 0.10198a,.;$
Then we apply the law of sines for triangle $BOC:quad$
$dfrac{OC}{sin (OBC)}=dfrac{BO}{sin (BCO)},;$ thus obtaning $;sin (OBC)=dfrac{OC}{BO}sin (BCO)approx0.039773,.;,$Finally we arrive at $angle OBC=sin^{-1}0.039773approx2.279^circ.;$ ($;$last digit is not precise, and it can be said
that $angle OBCapprox2.28^circ.)$ That's it. The problem is fully solved.
$$quad$$
$qquadqquadqquadqquadqquadqquadqquadqquad$* * * * * * * * * * * *$qquadqquadqquadqquadqquadqquadqquad$
$$quad$$
Now let's retrace all the steps and condense the whole computation into a single formula:
$$sin (OBC)=frac{OC cdot sin (BCO)}{sqrt{OC^2+BC^2-2cdot OC cdot BCcdot cos (BCO)}}=$$
$$=dfrac{acdotdfrac{sin (OAC)}{sin (AOC)}sin (BCO)}{sqrt{bigg(acdot dfrac{sin (OAC)}{sin (AOC)}bigg)^2+bigg(dfrac{a}{2cos(ACB)}bigg)^2-;2acdot dfrac{sin (OAC)}{sin (AOC)}cdot dfrac{a}{2cos(ACB)}cdot cos (BCO)}}$$
$qquadqquadqquadqquadqquadqquad$ Now we denote the angles with Greek letters
$qquadqquadqquadqquadqquadqquadqquadqquadangle ACB=alphaqquad(approx 6.6670^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquadqquadangle BCO=betaqquad(approx 0.5788^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquadqquadangle AOC=gammaqquad(approx 169.8570^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquadqquadangle OAC=phiqquad(approx 4.0548^circ)qquadqquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquad$and our expression becomes neat:
$$sin (OBC)=dfrac{acdotdfrac{sin phi}{sin gamma}sin beta}{sqrt{bigg(acdot dfrac{sin phi}{sin gamma}bigg)^2+bigg(dfrac{a}{2cosalpha}bigg)^2-;2acdot dfrac{sin phi}{sin gamma}cdot dfrac{a}{2cosalpha}cdot cos beta}}$$
This expression can be simplified by cancelling out $a$ and by dividing the numerator and denominator by the fraction $dfrac{sin phi}{sin gamma}$:
$$sin (OBC)=dfrac{sin beta}{sqrt{1+bigg(dfrac{sin gamma}{2,sin phi ,cosalpha},bigg)^2-;2cdot dfrac{sin gamma}{2,sin phi, cosalpha},cos beta}}=$$
$qquadqquadqquadqquad$ and now completing the square in the denominator
$$=dfrac{sin beta}{sqrt{bigg(dfrac{sin gamma}{2,sin phi ,cosalpha},bigg)^2-;2cdot dfrac{sin gamma}{2,sin phi, cosalpha},cos beta , + , cos^2{beta} - , cos^2{beta} , + , 1}}=$$
$qquadqquadqquadqquadqquadqquad$ remembering that $1-cos^2 beta=sin^2 beta$
$$=dfrac{sin beta}{sqrt{bigg(dfrac{sin gamma}{2,sin phi ,cosalpha}-cos{beta} bigg)^2+;sin^2beta}}=$$
$;$and then dividing the numerator and denominator by $,sin beta;$ we derive the final version of the formula
$$=dfrac{1}{sqrt{bigg(dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta} bigg)^2+;1}}=$$
$qquadqquadqquadqquad$ or $$angle OBC= sin^{-1}dfrac{1}{sqrt{bigg(dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta} bigg)^2+;1}}$$
Plugging in the angle values we once again obtain the same approximate result:
$$angle OBCapproxsin^{-1}frac{1}{sqrt{632.16}}approxsin^{-1}{0.039773}approx2.279^circ,.$$
$angle ACB=alphaqquad(approx 6.6670^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle BCO=betaqquad(approx 0.5788^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle AOC=gammaqquad(approx 169.8570^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle OAC=phiqquad(approx 4.0548^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
This radical in fact can be further simplified using the formula
$$sin {(cot^{-1}{x})}=frac{1}{sqrt{x^2+1}}:$$
$qquadqquadqquadqquadqquad$ here $;x=dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta}$
$$angle OBC=sin^{-1}{dfrac{1}{sqrt{bigg(dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta} bigg)^2+;1}}}=$$
$$sin^{-1}{frac{1}{sqrt{x^2+1}}}=sin^{-1}{(sin{(cot^{-1}{x})})}=cot^{-1}{x}=cot^{-1}{bigg(frac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta}bigg)}$$
This gives the result even faster. You get the same number (can check with a calculator) $approx2.279^circ,$
I've just turned the solution in my head through Ceva's theorem and it went without the radical (square root) i.e. you get the cotangent from a trigonometric equation resulting through the use of Ceva's theorem. This answer needs to be expanded to encompass Ceva's theorem, I guess. There's the beauty in all these interconnections. That's all. I need to add the second solution through Ceva's theorem.
$$quad$$
$qquadqquadqquadqquadqquadqquadqquadqquad$* * * * * * * * * * * *$qquadqquadqquadqquadqquadqquadqquad$
$$quad$$
Alternative solution using Ceva's theorem.
Let's write Ceva's theorem in trigonometric form:
$$frac{sin(ABO)}{sin(OBC)},frac{sin(OAC)}{sin(BAO)},frac{sin(BCO)}{sin(OCA)}=1$$
$qquadqquadqquadqquadqquadqquadangle ACB=alphaqquad(approx 6.6670^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadangle BCO=betaqquad(approx 0.5788^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadangle AOC=gammaqquad(approx 169.8570^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadangle OAC=phiqquad(approx 4.0548^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquadqquad$
We use the same denotations here but there are "more" angles now. Just like was shown in the very beginning most of these angles are obvious.
$;angle OAB=alpha-phi,,;angle OCA=alpha-beta,,;$and denoting $;angle OBC;$ (the angle we have to find) as $;x;$ it follows that $;angle ABO+angle OBC=180-2alpha;$ or $;angle ABO=180-2alpha-x,.$
Now we re-write Ceva's theorem as follows
$$frac{sin(180-2alpha-x)}{sin x},frac{sin phi}{sin(alpha-phi)},frac{sin beta}{sin(alpha-beta)}=1$$
Now we simplify the first factor of this equation:
$$frac{sin(180-2alpha-x)}{sin x}=frac{sin(2alpha+x)}{sin x}=frac{sin(2alpha)cos x + cos(2alpha)sin x}{sin x}=cot x sin (2alpha)+cos (2alpha)$$
So we have
$$[cot x,sin (2alpha)+cos (2alpha)],frac{sin phi}{sin(alpha-phi)},frac{sin beta}{sin(alpha-beta)}=1$$
$$cot x,sin (2alpha)+cos (2alpha)=,frac{sin(alpha-phi),sin(alpha-beta)}{sin phi ,sin beta}$$
$$cot x =,frac{sin(alpha-phi),sin(alpha-beta)}{sin(2alpha) sin phi ,sin beta}-cot{(2alpha)}$$
$$x =,cot^{-1}{bigg(,frac{sin(alpha-phi),sin(alpha-beta)}{sin(2alpha) sin phi ,sin beta}-cot{(2alpha)}bigg)}$$
Once again by plugging the numbers we get this value:
$$xapproxcot^{-1}{25.123}=tan^{-1}{frac{1}{25.123}}approx 2.279^circ.$$
Using Ceva's theorem gives us the solution through the inverse cotangent after solving the trivial trigonometric equation. So we don't have the radical here as opposed to when we used the law of cosines.
answered Dec 10 '18 at 18:12
Ken DracoKen Draco
56929
$endgroup$
$begingroup$
Regarding your PS: Ceva's Theorem is about the concurrency of lines through the vertices of a triangle, so it has something to do with this problem. In the "classical" presentation, Ceva's concurrency condition involves ratios of lengths of segments determined where the lines cut the opposite sides; those particular ratios are irrelevant here. However, the trigonometric form gives a condition involving the ratios of sines of angles formed by those lines; those ratios are not irrelevant here.
$endgroup$
– Blue
Dec 10 '18 at 18:21
$begingroup$
@Blue Math is one whole. So everything has to do with everything. You can zigzag this problem through a lot of ways. There's no doubt about that. Ceva's theorem is a beautiful one but I see no reason whatsoever trying to drag it through this specific problem.
$endgroup$
– Ken Draco
Dec 10 '18 at 18:27
1
$begingroup$
We're given information that lets us easily determine four of the six angles formed by the lines (namely, the angles at $A$ and $C$); and we have the sum of the two remaining angles (at $B$). Trigonometric Ceva then tells us the ratio of the sines of those $B$-angles, from which we can deduce the angles themselves with just a little work. There's not much (any?) dragging here; it's kinda what Ceva was meant to do. :)
$endgroup$
– Blue
Dec 10 '18 at 18:38
1
$begingroup$
@ Blue, Very good point. Thanks. Yes, indeed. I suspected that simplification was possible. doing it through Ceva reminded me about inverse trig formulas that extend into complex variables, and I left it with a radical. Gotta remedy that at least. And yes through Ceva it looks to be just as fast or probably faster.
$endgroup$
– Ken Draco
Dec 10 '18 at 19:53
1
$begingroup$
Well, Where is the error in my solution?? math.stackexchange.com/q/3031398/548054
$endgroup$
– Beginner
Dec 11 '18 at 20:08
|
show 2 more comments
$begingroup$
Let's denote $AC = a,.,$ Since the triangle $ABC$ is isosceles
$angle BAC=angle ACB=0.5cdot(180-angle ABC) approx6.6670^circ.qquadqquadqquadqquadqquad$
$angle OCA approx6.0882^circ.qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle OACapprox4.0548^circ.qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle BCO=angle ACB-angle OCAapprox 0.5788^circ.qquadqquadqquadqquadqquadqquadqquadqquad$
$angle AOC=180^circ-angle OAC-angle OCAapprox 169.8570^circ.$
We keep an extra digit for maximum precision.
$BC=dfrac{a}{2cos(ACB)}approx0.5034a;;$because triangle $ABC$ is isosceles.$;$ Applying the law of sines for the triangle $AOC$ we get $dfrac{OC}{sin (OAC)}=dfrac{AC}{sin (AOC)},.quad$ Hence $;OC=acdotdfrac{sin (OAC)}{sin (AOC)}approx0.40152a,.;$
Using the law of cosines for $triangle BOC,$:$quad BO=sqrt{OC^2+BC^2-2cdot OC cdot BCcdot cos(BCO)}approx$
$approx 0.10198a,.;$
Then we apply the law of sines for triangle $BOC:quad$
$dfrac{OC}{sin (OBC)}=dfrac{BO}{sin (BCO)},;$ thus obtaning $;sin (OBC)=dfrac{OC}{BO}sin (BCO)approx0.039773,.;,$Finally we arrive at $angle OBC=sin^{-1}0.039773approx2.279^circ.;$ ($;$last digit is not precise, and it can be said
that $angle OBCapprox2.28^circ.)$ That's it. The problem is fully solved.
$$quad$$
$qquadqquadqquadqquadqquadqquadqquadqquad$* * * * * * * * * * * *$qquadqquadqquadqquadqquadqquadqquad$
$$quad$$
Now let's retrace all the steps and condense the whole computation into a single formula:
$$sin (OBC)=frac{OC cdot sin (BCO)}{sqrt{OC^2+BC^2-2cdot OC cdot BCcdot cos (BCO)}}=$$
$$=dfrac{acdotdfrac{sin (OAC)}{sin (AOC)}sin (BCO)}{sqrt{bigg(acdot dfrac{sin (OAC)}{sin (AOC)}bigg)^2+bigg(dfrac{a}{2cos(ACB)}bigg)^2-;2acdot dfrac{sin (OAC)}{sin (AOC)}cdot dfrac{a}{2cos(ACB)}cdot cos (BCO)}}$$
$qquadqquadqquadqquadqquadqquad$ Now we denote the angles with Greek letters
$qquadqquadqquadqquadqquadqquadqquadqquadangle ACB=alphaqquad(approx 6.6670^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquadqquadangle BCO=betaqquad(approx 0.5788^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquadqquadangle AOC=gammaqquad(approx 169.8570^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquadqquadangle OAC=phiqquad(approx 4.0548^circ)qquadqquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquad$and our expression becomes neat:
$$sin (OBC)=dfrac{acdotdfrac{sin phi}{sin gamma}sin beta}{sqrt{bigg(acdot dfrac{sin phi}{sin gamma}bigg)^2+bigg(dfrac{a}{2cosalpha}bigg)^2-;2acdot dfrac{sin phi}{sin gamma}cdot dfrac{a}{2cosalpha}cdot cos beta}}$$
This expression can be simplified by cancelling out $a$ and by dividing the numerator and denominator by the fraction $dfrac{sin phi}{sin gamma}$:
$$sin (OBC)=dfrac{sin beta}{sqrt{1+bigg(dfrac{sin gamma}{2,sin phi ,cosalpha},bigg)^2-;2cdot dfrac{sin gamma}{2,sin phi, cosalpha},cos beta}}=$$
$qquadqquadqquadqquad$ and now completing the square in the denominator
$$=dfrac{sin beta}{sqrt{bigg(dfrac{sin gamma}{2,sin phi ,cosalpha},bigg)^2-;2cdot dfrac{sin gamma}{2,sin phi, cosalpha},cos beta , + , cos^2{beta} - , cos^2{beta} , + , 1}}=$$
$qquadqquadqquadqquadqquadqquad$ remembering that $1-cos^2 beta=sin^2 beta$
$$=dfrac{sin beta}{sqrt{bigg(dfrac{sin gamma}{2,sin phi ,cosalpha}-cos{beta} bigg)^2+;sin^2beta}}=$$
$;$and then dividing the numerator and denominator by $,sin beta;$ we derive the final version of the formula
$$=dfrac{1}{sqrt{bigg(dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta} bigg)^2+;1}}=$$
$qquadqquadqquadqquad$ or $$angle OBC= sin^{-1}dfrac{1}{sqrt{bigg(dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta} bigg)^2+;1}}$$
Plugging in the angle values we once again obtain the same approximate result:
$$angle OBCapproxsin^{-1}frac{1}{sqrt{632.16}}approxsin^{-1}{0.039773}approx2.279^circ,.$$
$angle ACB=alphaqquad(approx 6.6670^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle BCO=betaqquad(approx 0.5788^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle AOC=gammaqquad(approx 169.8570^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle OAC=phiqquad(approx 4.0548^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
This radical in fact can be further simplified using the formula
$$sin {(cot^{-1}{x})}=frac{1}{sqrt{x^2+1}}:$$
$qquadqquadqquadqquadqquad$ here $;x=dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta}$
$$angle OBC=sin^{-1}{dfrac{1}{sqrt{bigg(dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta} bigg)^2+;1}}}=$$
$$sin^{-1}{frac{1}{sqrt{x^2+1}}}=sin^{-1}{(sin{(cot^{-1}{x})})}=cot^{-1}{x}=cot^{-1}{bigg(frac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta}bigg)}$$
This gives the result even faster. You get the same number (can check with a calculator) $approx2.279^circ,$
I've just turned the solution in my head through Ceva's theorem and it went without the radical (square root) i.e. you get the cotangent from a trigonometric equation resulting through the use of Ceva's theorem. This answer needs to be expanded to encompass Ceva's theorem, I guess. There's the beauty in all these interconnections. That's all. I need to add the second solution through Ceva's theorem.
$$quad$$
$qquadqquadqquadqquadqquadqquadqquadqquad$* * * * * * * * * * * *$qquadqquadqquadqquadqquadqquadqquad$
$$quad$$
Alternative solution using Ceva's theorem.
Let's write Ceva's theorem in trigonometric form:
$$frac{sin(ABO)}{sin(OBC)},frac{sin(OAC)}{sin(BAO)},frac{sin(BCO)}{sin(OCA)}=1$$
$qquadqquadqquadqquadqquadqquadangle ACB=alphaqquad(approx 6.6670^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadangle BCO=betaqquad(approx 0.5788^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadangle AOC=gammaqquad(approx 169.8570^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadangle OAC=phiqquad(approx 4.0548^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquadqquad$
We use the same denotations here but there are "more" angles now. Just like was shown in the very beginning most of these angles are obvious.
$;angle OAB=alpha-phi,,;angle OCA=alpha-beta,,;$and denoting $;angle OBC;$ (the angle we have to find) as $;x;$ it follows that $;angle ABO+angle OBC=180-2alpha;$ or $;angle ABO=180-2alpha-x,.$
Now we re-write Ceva's theorem as follows
$$frac{sin(180-2alpha-x)}{sin x},frac{sin phi}{sin(alpha-phi)},frac{sin beta}{sin(alpha-beta)}=1$$
Now we simplify the first factor of this equation:
$$frac{sin(180-2alpha-x)}{sin x}=frac{sin(2alpha+x)}{sin x}=frac{sin(2alpha)cos x + cos(2alpha)sin x}{sin x}=cot x sin (2alpha)+cos (2alpha)$$
So we have
$$[cot x,sin (2alpha)+cos (2alpha)],frac{sin phi}{sin(alpha-phi)},frac{sin beta}{sin(alpha-beta)}=1$$
$$cot x,sin (2alpha)+cos (2alpha)=,frac{sin(alpha-phi),sin(alpha-beta)}{sin phi ,sin beta}$$
$$cot x =,frac{sin(alpha-phi),sin(alpha-beta)}{sin(2alpha) sin phi ,sin beta}-cot{(2alpha)}$$
$$x =,cot^{-1}{bigg(,frac{sin(alpha-phi),sin(alpha-beta)}{sin(2alpha) sin phi ,sin beta}-cot{(2alpha)}bigg)}$$
Once again by plugging the numbers we get this value:
$$xapproxcot^{-1}{25.123}=tan^{-1}{frac{1}{25.123}}approx 2.279^circ.$$
Using Ceva's theorem gives us the solution through the inverse cotangent after solving the trivial trigonometric equation. So we don't have the radical here as opposed to when we used the law of cosines.
answered Dec 10 '18 at 18:12
Ken DracoKen Draco
56929
$endgroup$
Let's denote $AC = a,.,$ Since the triangle $ABC$ is isosceles
$angle BAC=angle ACB=0.5cdot(180-angle ABC) approx6.6670^circ.qquadqquadqquadqquadqquad$
$angle OCA approx6.0882^circ.qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle OACapprox4.0548^circ.qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle BCO=angle ACB-angle OCAapprox 0.5788^circ.qquadqquadqquadqquadqquadqquadqquadqquad$
$angle AOC=180^circ-angle OAC-angle OCAapprox 169.8570^circ.$
We keep an extra digit for maximum precision.
$BC=dfrac{a}{2cos(ACB)}approx0.5034a;;$because triangle $ABC$ is isosceles.$;$ Applying the law of sines for the triangle $AOC$ we get $dfrac{OC}{sin (OAC)}=dfrac{AC}{sin (AOC)},.quad$ Hence $;OC=acdotdfrac{sin (OAC)}{sin (AOC)}approx0.40152a,.;$
Using the law of cosines for $triangle BOC,$:$quad BO=sqrt{OC^2+BC^2-2cdot OC cdot BCcdot cos(BCO)}approx$
$approx 0.10198a,.;$
Then we apply the law of sines for triangle $BOC:quad$
$dfrac{OC}{sin (OBC)}=dfrac{BO}{sin (BCO)},;$ thus obtaning $;sin (OBC)=dfrac{OC}{BO}sin (BCO)approx0.039773,.;,$Finally we arrive at $angle OBC=sin^{-1}0.039773approx2.279^circ.;$ ($;$last digit is not precise, and it can be said
that $angle OBCapprox2.28^circ.)$ That's it. The problem is fully solved.
$$quad$$
$qquadqquadqquadqquadqquadqquadqquadqquad$* * * * * * * * * * * *$qquadqquadqquadqquadqquadqquadqquad$
$$quad$$
Now let's retrace all the steps and condense the whole computation into a single formula:
$$sin (OBC)=frac{OC cdot sin (BCO)}{sqrt{OC^2+BC^2-2cdot OC cdot BCcdot cos (BCO)}}=$$
$$=dfrac{acdotdfrac{sin (OAC)}{sin (AOC)}sin (BCO)}{sqrt{bigg(acdot dfrac{sin (OAC)}{sin (AOC)}bigg)^2+bigg(dfrac{a}{2cos(ACB)}bigg)^2-;2acdot dfrac{sin (OAC)}{sin (AOC)}cdot dfrac{a}{2cos(ACB)}cdot cos (BCO)}}$$
$qquadqquadqquadqquadqquadqquad$ Now we denote the angles with Greek letters
$qquadqquadqquadqquadqquadqquadqquadqquadangle ACB=alphaqquad(approx 6.6670^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquadqquadangle BCO=betaqquad(approx 0.5788^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquadqquadangle AOC=gammaqquad(approx 169.8570^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquadqquadangle OAC=phiqquad(approx 4.0548^circ)qquadqquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadqquad$and our expression becomes neat:
$$sin (OBC)=dfrac{acdotdfrac{sin phi}{sin gamma}sin beta}{sqrt{bigg(acdot dfrac{sin phi}{sin gamma}bigg)^2+bigg(dfrac{a}{2cosalpha}bigg)^2-;2acdot dfrac{sin phi}{sin gamma}cdot dfrac{a}{2cosalpha}cdot cos beta}}$$
This expression can be simplified by cancelling out $a$ and by dividing the numerator and denominator by the fraction $dfrac{sin phi}{sin gamma}$:
$$sin (OBC)=dfrac{sin beta}{sqrt{1+bigg(dfrac{sin gamma}{2,sin phi ,cosalpha},bigg)^2-;2cdot dfrac{sin gamma}{2,sin phi, cosalpha},cos beta}}=$$
$qquadqquadqquadqquad$ and now completing the square in the denominator
$$=dfrac{sin beta}{sqrt{bigg(dfrac{sin gamma}{2,sin phi ,cosalpha},bigg)^2-;2cdot dfrac{sin gamma}{2,sin phi, cosalpha},cos beta , + , cos^2{beta} - , cos^2{beta} , + , 1}}=$$
$qquadqquadqquadqquadqquadqquad$ remembering that $1-cos^2 beta=sin^2 beta$
$$=dfrac{sin beta}{sqrt{bigg(dfrac{sin gamma}{2,sin phi ,cosalpha}-cos{beta} bigg)^2+;sin^2beta}}=$$
$;$and then dividing the numerator and denominator by $,sin beta;$ we derive the final version of the formula
$$=dfrac{1}{sqrt{bigg(dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta} bigg)^2+;1}}=$$
$qquadqquadqquadqquad$ or $$angle OBC= sin^{-1}dfrac{1}{sqrt{bigg(dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta} bigg)^2+;1}}$$
Plugging in the angle values we once again obtain the same approximate result:
$$angle OBCapproxsin^{-1}frac{1}{sqrt{632.16}}approxsin^{-1}{0.039773}approx2.279^circ,.$$
$angle ACB=alphaqquad(approx 6.6670^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle BCO=betaqquad(approx 0.5788^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle AOC=gammaqquad(approx 169.8570^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
$angle OAC=phiqquad(approx 4.0548^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquad$
This radical in fact can be further simplified using the formula
$$sin {(cot^{-1}{x})}=frac{1}{sqrt{x^2+1}}:$$
$qquadqquadqquadqquadqquad$ here $;x=dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta}$
$$angle OBC=sin^{-1}{dfrac{1}{sqrt{bigg(dfrac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta} bigg)^2+;1}}}=$$
$$sin^{-1}{frac{1}{sqrt{x^2+1}}}=sin^{-1}{(sin{(cot^{-1}{x})})}=cot^{-1}{x}=cot^{-1}{bigg(frac{sin gamma}{2,sin beta ,sin phi ,cosalpha}-cot{beta}bigg)}$$
This gives the result even faster. You get the same number (can check with a calculator) $approx2.279^circ,$
I've just turned the solution in my head through Ceva's theorem and it went without the radical (square root) i.e. you get the cotangent from a trigonometric equation resulting through the use of Ceva's theorem. This answer needs to be expanded to encompass Ceva's theorem, I guess. There's the beauty in all these interconnections. That's all. I need to add the second solution through Ceva's theorem.
$$quad$$
$qquadqquadqquadqquadqquadqquadqquadqquad$* * * * * * * * * * * *$qquadqquadqquadqquadqquadqquadqquad$
$$quad$$
Alternative solution using Ceva's theorem.
Let's write Ceva's theorem in trigonometric form:
$$frac{sin(ABO)}{sin(OBC)},frac{sin(OAC)}{sin(BAO)},frac{sin(BCO)}{sin(OCA)}=1$$
$qquadqquadqquadqquadqquadqquadangle ACB=alphaqquad(approx 6.6670^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadangle BCO=betaqquad(approx 0.5788^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadangle AOC=gammaqquad(approx 169.8570^circ)qquadqquadqquadqquad$
$qquadqquadqquadqquadqquadqquadangle OAC=phiqquad(approx 4.0548^circ)qquadqquadqquadqquadqquadqquadqquadqquadqquadqquad$
We use the same denotations here but there are "more" angles now. Just like was shown in the very beginning most of these angles are obvious.
$;angle OAB=alpha-phi,,;angle OCA=alpha-beta,,;$and denoting $;angle OBC;$ (the angle we have to find) as $;x;$ it follows that $;angle ABO+angle OBC=180-2alpha;$ or $;angle ABO=180-2alpha-x,.$
Now we re-write Ceva's theorem as follows
$$frac{sin(180-2alpha-x)}{sin x},frac{sin phi}{sin(alpha-phi)},frac{sin beta}{sin(alpha-beta)}=1$$
Now we simplify the first factor of this equation:
$$frac{sin(180-2alpha-x)}{sin x}=frac{sin(2alpha+x)}{sin x}=frac{sin(2alpha)cos x + cos(2alpha)sin x}{sin x}=cot x sin (2alpha)+cos (2alpha)$$
So we have
$$[cot x,sin (2alpha)+cos (2alpha)],frac{sin phi}{sin(alpha-phi)},frac{sin beta}{sin(alpha-beta)}=1$$
$$cot x,sin (2alpha)+cos (2alpha)=,frac{sin(alpha-phi),sin(alpha-beta)}{sin phi ,sin beta}$$
$$cot x =,frac{sin(alpha-phi),sin(alpha-beta)}{sin(2alpha) sin phi ,sin beta}-cot{(2alpha)}$$
$$x =,cot^{-1}{bigg(,frac{sin(alpha-phi),sin(alpha-beta)}{sin(2alpha) sin phi ,sin beta}-cot{(2alpha)}bigg)}$$
Once again by plugging the numbers we get this value:
$$xapproxcot^{-1}{25.123}=tan^{-1}{frac{1}{25.123}}approx 2.279^circ.$$
Using Ceva's theorem gives us the solution through the inverse cotangent after solving the trivial trigonometric equation. So we don't have the radical here as opposed to when we used the law of cosines.
answered Dec 10 '18 at 18:12
Ken DracoKen Draco
56929
edited Dec 11 '18 at 10:08
answered Dec 10 '18 at 18:12
Ken DracoKen Draco
56929
answered Dec 10 '18 at 18:12
Ken DracoKen Draco
56929
answered Dec 10 '18 at 18:12
Ken DracoKen Draco
56929
56929
$begingroup$
Regarding your PS: Ceva's Theorem is about the concurrency of lines through the vertices of a triangle, so it has something to do with this problem. In the "classical" presentation, Ceva's concurrency condition involves ratios of lengths of segments determined where the lines cut the opposite sides; those particular ratios are irrelevant here. However, the trigonometric form gives a condition involving the ratios of sines of angles formed by those lines; those ratios are not irrelevant here.
$endgroup$
– Blue
Dec 10 '18 at 18:21
$begingroup$
@Blue Math is one whole. So everything has to do with everything. You can zigzag this problem through a lot of ways. There's no doubt about that. Ceva's theorem is a beautiful one but I see no reason whatsoever trying to drag it through this specific problem.
$endgroup$
– Ken Draco
Dec 10 '18 at 18:27
1
$begingroup$
We're given information that lets us easily determine four of the six angles formed by the lines (namely, the angles at $A$ and $C$); and we have the sum of the two remaining angles (at $B$). Trigonometric Ceva then tells us the ratio of the sines of those $B$-angles, from which we can deduce the angles themselves with just a little work. There's not much (any?) dragging here; it's kinda what Ceva was meant to do. :)
$endgroup$
– Blue
Dec 10 '18 at 18:38
1
$begingroup$
@ Blue, Very good point. Thanks. Yes, indeed. I suspected that simplification was possible. doing it through Ceva reminded me about inverse trig formulas that extend into complex variables, and I left it with a radical. Gotta remedy that at least. And yes through Ceva it looks to be just as fast or probably faster.
$endgroup$
– Ken Draco
Dec 10 '18 at 19:53
1
$begingroup$
Well, Where is the error in my solution?? math.stackexchange.com/q/3031398/548054
$endgroup$
– Beginner
Dec 11 '18 at 20:08
|
show 2 more comments
$begingroup$
Regarding your PS: Ceva's Theorem is about the concurrency of lines through the vertices of a triangle, so it has something to do with this problem. In the "classical" presentation, Ceva's concurrency condition involves ratios of lengths of segments determined where the lines cut the opposite sides; those particular ratios are irrelevant here. However, the trigonometric form gives a condition involving the ratios of sines of angles formed by those lines; those ratios are not irrelevant here.
$endgroup$
– Blue
Dec 10 '18 at 18:21
$begingroup$
@Blue Math is one whole. So everything has to do with everything. You can zigzag this problem through a lot of ways. There's no doubt about that. Ceva's theorem is a beautiful one but I see no reason whatsoever trying to drag it through this specific problem.
$endgroup$
– Ken Draco
Dec 10 '18 at 18:27
1
$begingroup$
We're given information that lets us easily determine four of the six angles formed by the lines (namely, the angles at $A$ and $C$); and we have the sum of the two remaining angles (at $B$). Trigonometric Ceva then tells us the ratio of the sines of those $B$-angles, from which we can deduce the angles themselves with just a little work. There's not much (any?) dragging here; it's kinda what Ceva was meant to do. :)
$endgroup$
– Blue
Dec 10 '18 at 18:38
1
$begingroup$
@ Blue, Very good point. Thanks. Yes, indeed. I suspected that simplification was possible. doing it through Ceva reminded me about inverse trig formulas that extend into complex variables, and I left it with a radical. Gotta remedy that at least. And yes through Ceva it looks to be just as fast or probably faster.
$endgroup$
– Ken Draco
Dec 10 '18 at 19:53
1
$begingroup$
Well, Where is the error in my solution?? math.stackexchange.com/q/3031398/548054
$endgroup$
– Beginner
Dec 11 '18 at 20:08
$begingroup$
Regarding your PS: Ceva's Theorem is about the concurrency of lines through the vertices of a triangle, so it has something to do with this problem. In the "classical" presentation, Ceva's concurrency condition involves ratios of lengths of segments determined where the lines cut the opposite sides; those particular ratios are irrelevant here. However, the trigonometric form gives a condition involving the ratios of sines of angles formed by those lines; those ratios are not irrelevant here.
$endgroup$
– Blue
Dec 10 '18 at 18:21
$begingroup$
Regarding your PS: Ceva's Theorem is about the concurrency of lines through the vertices of a triangle, so it has something to do with this problem. In the "classical" presentation, Ceva's concurrency condition involves ratios of lengths of segments determined where the lines cut the opposite sides; those particular ratios are irrelevant here. However, the trigonometric form gives a condition involving the ratios of sines of angles formed by those lines; those ratios are not irrelevant here.
$endgroup$
– Blue
Dec 10 '18 at 18:21
$begingroup$
@Blue Math is one whole. So everything has to do with everything. You can zigzag this problem through a lot of ways. There's no doubt about that. Ceva's theorem is a beautiful one but I see no reason whatsoever trying to drag it through this specific problem.
$endgroup$
– Ken Draco
Dec 10 '18 at 18:27
$begingroup$
@Blue Math is one whole. So everything has to do with everything. You can zigzag this problem through a lot of ways. There's no doubt about that. Ceva's theorem is a beautiful one but I see no reason whatsoever trying to drag it through this specific problem.
$endgroup$
– Ken Draco
Dec 10 '18 at 18:27
1
1
$begingroup$
We're given information that lets us easily determine four of the six angles formed by the lines (namely, the angles at $A$ and $C$); and we have the sum of the two remaining angles (at $B$). Trigonometric Ceva then tells us the ratio of the sines of those $B$-angles, from which we can deduce the angles themselves with just a little work. There's not much (any?) dragging here; it's kinda what Ceva was meant to do. :)
$endgroup$
– Blue
Dec 10 '18 at 18:38
$begingroup$
We're given information that lets us easily determine four of the six angles formed by the lines (namely, the angles at $A$ and $C$); and we have the sum of the two remaining angles (at $B$). Trigonometric Ceva then tells us the ratio of the sines of those $B$-angles, from which we can deduce the angles themselves with just a little work. There's not much (any?) dragging here; it's kinda what Ceva was meant to do. :)
$endgroup$
– Blue
Dec 10 '18 at 18:38
1
1
$begingroup$
@ Blue, Very good point. Thanks. Yes, indeed. I suspected that simplification was possible. doing it through Ceva reminded me about inverse trig formulas that extend into complex variables, and I left it with a radical. Gotta remedy that at least. And yes through Ceva it looks to be just as fast or probably faster.
$endgroup$
– Ken Draco
Dec 10 '18 at 19:53
$begingroup$
@ Blue, Very good point. Thanks. Yes, indeed. I suspected that simplification was possible. doing it through Ceva reminded me about inverse trig formulas that extend into complex variables, and I left it with a radical. Gotta remedy that at least. And yes through Ceva it looks to be just as fast or probably faster.
$endgroup$
– Ken Draco
Dec 10 '18 at 19:53
1
1
$begingroup$
Well, Where is the error in my solution?? math.stackexchange.com/q/3031398/548054
$endgroup$
– Beginner
Dec 11 '18 at 20:08
$begingroup$
Well, Where is the error in my solution?? math.stackexchange.com/q/3031398/548054
$endgroup$
– Beginner
Dec 11 '18 at 20:08
|
show 2 more comments
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$begingroup$
Are the commas decimal points or thousands separators?
$endgroup$
– saulspatz
Dec 2 '18 at 21:42
$begingroup$
@saulspatz decimal points of course
$endgroup$
– Beginner
Dec 2 '18 at 21:46
2
$begingroup$
I don't think the answer can be expressed in a closed form. Apparently, you have to solve for $x:=angle OBC$ from $$frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)},,$$ where $beta:=angle ABC$, $alpha:=angle OAC$, $gamma:=angle OCA$, and $theta:=angle BAC=angle ACB$. Mathematica says that $$xapprox 0.033921 approx 1.94353^circ,.$$ The given parameters are so arbitrary that I don't see any reason we should expect a nice answer.
$endgroup$
– Batominovski
Dec 2 '18 at 21:50
2
$begingroup$
However, maybe the point of this question is to illustrate how to use Ceva's Theorem in the trigonometric form.
$endgroup$
– Batominovski
Dec 2 '18 at 21:57
1
$begingroup$
Let's just say that the theorem is basically a derivative of the Law of Sines. And Google is your friend.
$endgroup$
– Batominovski
Dec 2 '18 at 22:09