How do we evaluate the degree of $x$ using sine law?
$begingroup$
Given that
ABC is a triangle, $|EC| = |BC| = |BD| $, $angle CBA= 80^circ,angle ACB= 60^circ, angle EDA= x^circ $
Evaluate $x$
I want to solve this for $x$ using law of sines if possible.
My attempt:
From the property of triangle, the sum of the angles will be equal to $180$.
$$angle BAC = 180 - 80 - 60 = 40^circ $$
In $triangle ABC$,
$$frac{sin 40}{|BC|} = frac{sin 80}{|AC|} implies frac{|AC|}{|BC|} = frac{sin 80}{sin40}$$
Could you help me take it from there?
Regards
geometry trigonometry
asked Nov 30 '18 at 12:11
HamiltonHamilton
1798
$endgroup$
add a comment |
$begingroup$
Given that
ABC is a triangle, $|EC| = |BC| = |BD| $, $angle CBA= 80^circ,angle ACB= 60^circ, angle EDA= x^circ $
Evaluate $x$
I want to solve this for $x$ using law of sines if possible.
My attempt:
From the property of triangle, the sum of the angles will be equal to $180$.
$$angle BAC = 180 - 80 - 60 = 40^circ $$
In $triangle ABC$,
$$frac{sin 40}{|BC|} = frac{sin 80}{|AC|} implies frac{|AC|}{|BC|} = frac{sin 80}{sin40}$$
Could you help me take it from there?
Regards
geometry trigonometry
asked Nov 30 '18 at 12:11
HamiltonHamilton
1798
$endgroup$
add a comment |
$begingroup$
Given that
ABC is a triangle, $|EC| = |BC| = |BD| $, $angle CBA= 80^circ,angle ACB= 60^circ, angle EDA= x^circ $
Evaluate $x$
I want to solve this for $x$ using law of sines if possible.
My attempt:
From the property of triangle, the sum of the angles will be equal to $180$.
$$angle BAC = 180 - 80 - 60 = 40^circ $$
In $triangle ABC$,
$$frac{sin 40}{|BC|} = frac{sin 80}{|AC|} implies frac{|AC|}{|BC|} = frac{sin 80}{sin40}$$
Could you help me take it from there?
Regards
geometry trigonometry
asked Nov 30 '18 at 12:11
HamiltonHamilton
1798
$endgroup$
Given that
ABC is a triangle, $|EC| = |BC| = |BD| $, $angle CBA= 80^circ,angle ACB= 60^circ, angle EDA= x^circ $
Evaluate $x$
I want to solve this for $x$ using law of sines if possible.
My attempt:
From the property of triangle, the sum of the angles will be equal to $180$.
$$angle BAC = 180 - 80 - 60 = 40^circ $$
In $triangle ABC$,
$$frac{sin 40}{|BC|} = frac{sin 80}{|AC|} implies frac{|AC|}{|BC|} = frac{sin 80}{sin40}$$
Could you help me take it from there?
Regards
geometry trigonometry
geometry trigonometry
asked Nov 30 '18 at 12:11
HamiltonHamilton
1798
asked Nov 30 '18 at 12:11
HamiltonHamilton
1798
asked Nov 30 '18 at 12:11
HamiltonHamilton
1798
asked Nov 30 '18 at 12:11
HamiltonHamilton
1798
asked Nov 30 '18 at 12:11
HamiltonHamilton
1798
1798
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You don't need trigonometry to solve for $x$. Observe that $BCE$ is an equilateral triangle, so $BE=BC$, so $BDE$ is an isosceles triangle. From this, it's easy to see that $x=100°$.
answered Nov 30 '18 at 12:20
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.7k72445
$endgroup$
add a comment |
$begingroup$
Since $|BC|=|EC|$ and $angle BCE=60$ then $triangle BCE$ is equilateral.
Hence $|BE|=|BC|=|BD|$, so that $triangle BDE$ is isosceles with $angle DBE=80°-60°=20°$.
Hence, $angle BDE=(180°-20°)/2=80°$ so that $angle ADE=180°-80°=100°$.
answered Nov 30 '18 at 12:33
AntinousAntinous
5,71042051
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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$begingroup$
You don't need trigonometry to solve for $x$. Observe that $BCE$ is an equilateral triangle, so $BE=BC$, so $BDE$ is an isosceles triangle. From this, it's easy to see that $x=100°$.
answered Nov 30 '18 at 12:20
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.7k72445
$endgroup$
add a comment |
$begingroup$
You don't need trigonometry to solve for $x$. Observe that $BCE$ is an equilateral triangle, so $BE=BC$, so $BDE$ is an isosceles triangle. From this, it's easy to see that $x=100°$.
answered Nov 30 '18 at 12:20
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.7k72445
$endgroup$
add a comment |
$begingroup$
You don't need trigonometry to solve for $x$. Observe that $BCE$ is an equilateral triangle, so $BE=BC$, so $BDE$ is an isosceles triangle. From this, it's easy to see that $x=100°$.
answered Nov 30 '18 at 12:20
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.7k72445
$endgroup$
You don't need trigonometry to solve for $x$. Observe that $BCE$ is an equilateral triangle, so $BE=BC$, so $BDE$ is an isosceles triangle. From this, it's easy to see that $x=100°$.
answered Nov 30 '18 at 12:20
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.7k72445
edited Nov 30 '18 at 13:46
answered Nov 30 '18 at 12:20
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.7k72445
answered Nov 30 '18 at 12:20
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.7k72445
answered Nov 30 '18 at 12:20
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.7k72445
12.7k72445
add a comment |
add a comment |
$begingroup$
Since $|BC|=|EC|$ and $angle BCE=60$ then $triangle BCE$ is equilateral.
Hence $|BE|=|BC|=|BD|$, so that $triangle BDE$ is isosceles with $angle DBE=80°-60°=20°$.
Hence, $angle BDE=(180°-20°)/2=80°$ so that $angle ADE=180°-80°=100°$.
answered Nov 30 '18 at 12:33
AntinousAntinous
5,71042051
$endgroup$
add a comment |
$begingroup$
Since $|BC|=|EC|$ and $angle BCE=60$ then $triangle BCE$ is equilateral.
Hence $|BE|=|BC|=|BD|$, so that $triangle BDE$ is isosceles with $angle DBE=80°-60°=20°$.
Hence, $angle BDE=(180°-20°)/2=80°$ so that $angle ADE=180°-80°=100°$.
answered Nov 30 '18 at 12:33
AntinousAntinous
5,71042051
$endgroup$
add a comment |
$begingroup$
Since $|BC|=|EC|$ and $angle BCE=60$ then $triangle BCE$ is equilateral.
Hence $|BE|=|BC|=|BD|$, so that $triangle BDE$ is isosceles with $angle DBE=80°-60°=20°$.
Hence, $angle BDE=(180°-20°)/2=80°$ so that $angle ADE=180°-80°=100°$.
answered Nov 30 '18 at 12:33
AntinousAntinous
5,71042051
$endgroup$
Since $|BC|=|EC|$ and $angle BCE=60$ then $triangle BCE$ is equilateral.
Hence $|BE|=|BC|=|BD|$, so that $triangle BDE$ is isosceles with $angle DBE=80°-60°=20°$.
Hence, $angle BDE=(180°-20°)/2=80°$ so that $angle ADE=180°-80°=100°$.
answered Nov 30 '18 at 12:33
AntinousAntinous
5,71042051
answered Nov 30 '18 at 12:33
AntinousAntinous
5,71042051
answered Nov 30 '18 at 12:33
AntinousAntinous
5,71042051
answered Nov 30 '18 at 12:33
AntinousAntinous
5,71042051
5,71042051
add a comment |
add a comment |
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