Finding the location of points of a triangle given the angle and length ratio.
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Given that a point P is located at (-2.5,4.33) I need to locate the points A and B such that $frac{PA}{PB} = frac{4.77}{8}$ and $angle APB = 55^o $.
A and B must be on the -ve part of the x axis.
I need to get A and B located in order to solve a problem in Control Systems Lag-Lead Compensator design. The problem is found in Modern Control Engineering by Ogata.
geometry triangle angle
asked Dec 2 '18 at 9:39
Hasan HammoudHasan Hammoud
254
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add a comment |
$begingroup$
Given that a point P is located at (-2.5,4.33) I need to locate the points A and B such that $frac{PA}{PB} = frac{4.77}{8}$ and $angle APB = 55^o $.
A and B must be on the -ve part of the x axis.
I need to get A and B located in order to solve a problem in Control Systems Lag-Lead Compensator design. The problem is found in Modern Control Engineering by Ogata.
geometry triangle angle
asked Dec 2 '18 at 9:39
Hasan HammoudHasan Hammoud
254
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1
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Isn't there any other condition or information? The problem as written has infinity solutions (A,B)
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– user376343
Dec 2 '18 at 9:45
1
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That information alone is not enough. You can rotate the triangle around point $P$ however you like and get infinitely many different values for $A$ and $B$. Or are you looking for a general expression?
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– platty
Dec 2 '18 at 9:46
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Sorry my bad you both are right I forgot to mention that A and B must be on the x-axis (-ve part of x-axis to be specific)
$endgroup$
– Hasan Hammoud
Dec 2 '18 at 9:48
add a comment |
$begingroup$
Given that a point P is located at (-2.5,4.33) I need to locate the points A and B such that $frac{PA}{PB} = frac{4.77}{8}$ and $angle APB = 55^o $.
A and B must be on the -ve part of the x axis.
I need to get A and B located in order to solve a problem in Control Systems Lag-Lead Compensator design. The problem is found in Modern Control Engineering by Ogata.
geometry triangle angle
asked Dec 2 '18 at 9:39
Hasan HammoudHasan Hammoud
254
$endgroup$
Given that a point P is located at (-2.5,4.33) I need to locate the points A and B such that $frac{PA}{PB} = frac{4.77}{8}$ and $angle APB = 55^o $.
A and B must be on the -ve part of the x axis.
I need to get A and B located in order to solve a problem in Control Systems Lag-Lead Compensator design. The problem is found in Modern Control Engineering by Ogata.
geometry triangle angle
geometry triangle angle
asked Dec 2 '18 at 9:39
Hasan HammoudHasan Hammoud
254
asked Dec 2 '18 at 9:39
Hasan HammoudHasan Hammoud
254
edited Dec 2 '18 at 9:48
Hasan Hammoud
asked Dec 2 '18 at 9:39
Hasan HammoudHasan Hammoud
254
asked Dec 2 '18 at 9:39
Hasan HammoudHasan Hammoud
254
asked Dec 2 '18 at 9:39
Hasan HammoudHasan Hammoud
254
254
1
$begingroup$
Isn't there any other condition or information? The problem as written has infinity solutions (A,B)
$endgroup$
– user376343
Dec 2 '18 at 9:45
1
$begingroup$
That information alone is not enough. You can rotate the triangle around point $P$ however you like and get infinitely many different values for $A$ and $B$. Or are you looking for a general expression?
$endgroup$
– platty
Dec 2 '18 at 9:46
$begingroup$
Sorry my bad you both are right I forgot to mention that A and B must be on the x-axis (-ve part of x-axis to be specific)
$endgroup$
– Hasan Hammoud
Dec 2 '18 at 9:48
add a comment |
1
$begingroup$
Isn't there any other condition or information? The problem as written has infinity solutions (A,B)
$endgroup$
– user376343
Dec 2 '18 at 9:45
1
$begingroup$
That information alone is not enough. You can rotate the triangle around point $P$ however you like and get infinitely many different values for $A$ and $B$. Or are you looking for a general expression?
$endgroup$
– platty
Dec 2 '18 at 9:46
$begingroup$
Sorry my bad you both are right I forgot to mention that A and B must be on the x-axis (-ve part of x-axis to be specific)
$endgroup$
– Hasan Hammoud
Dec 2 '18 at 9:48
1
1
$begingroup$
Isn't there any other condition or information? The problem as written has infinity solutions (A,B)
$endgroup$
– user376343
Dec 2 '18 at 9:45
$begingroup$
Isn't there any other condition or information? The problem as written has infinity solutions (A,B)
$endgroup$
– user376343
Dec 2 '18 at 9:45
1
1
$begingroup$
That information alone is not enough. You can rotate the triangle around point $P$ however you like and get infinitely many different values for $A$ and $B$. Or are you looking for a general expression?
$endgroup$
– platty
Dec 2 '18 at 9:46
$begingroup$
That information alone is not enough. You can rotate the triangle around point $P$ however you like and get infinitely many different values for $A$ and $B$. Or are you looking for a general expression?
$endgroup$
– platty
Dec 2 '18 at 9:46
$begingroup$
Sorry my bad you both are right I forgot to mention that A and B must be on the x-axis (-ve part of x-axis to be specific)
$endgroup$
– Hasan Hammoud
Dec 2 '18 at 9:48
$begingroup$
Sorry my bad you both are right I forgot to mention that A and B must be on the x-axis (-ve part of x-axis to be specific)
$endgroup$
– Hasan Hammoud
Dec 2 '18 at 9:48
add a comment |
1 Answer
1
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votes
$begingroup$
HINT: Let $A= (x_a , 0)$ and $B=(x_b, 0)$, with $x_a < x_b < 0$. You need to find $x_a$ and $x_b$. Consider a system of two equations:
In the first you write down the condition regarding the lenght ratio.
The second equation can be written using the Law of cosine, (see https://en.wikipedia.org/wiki/Law_of_cosines), since you know the cosine of the angle $ angle APB$.
answered Dec 2 '18 at 10:16
HermioneHermione
19619
$endgroup$
$begingroup$
Thanks for your answer thats exactly what I did however the equations got very messy. I was thinking if there is a more mathematically easier way of getting this solved
$endgroup$
– Hasan Hammoud
Dec 2 '18 at 10:21
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT: Let $A= (x_a , 0)$ and $B=(x_b, 0)$, with $x_a < x_b < 0$. You need to find $x_a$ and $x_b$. Consider a system of two equations:
In the first you write down the condition regarding the lenght ratio.
The second equation can be written using the Law of cosine, (see https://en.wikipedia.org/wiki/Law_of_cosines), since you know the cosine of the angle $ angle APB$.
answered Dec 2 '18 at 10:16
HermioneHermione
19619
$endgroup$
$begingroup$
Thanks for your answer thats exactly what I did however the equations got very messy. I was thinking if there is a more mathematically easier way of getting this solved
$endgroup$
– Hasan Hammoud
Dec 2 '18 at 10:21
add a comment |
$begingroup$
HINT: Let $A= (x_a , 0)$ and $B=(x_b, 0)$, with $x_a < x_b < 0$. You need to find $x_a$ and $x_b$. Consider a system of two equations:
In the first you write down the condition regarding the lenght ratio.
The second equation can be written using the Law of cosine, (see https://en.wikipedia.org/wiki/Law_of_cosines), since you know the cosine of the angle $ angle APB$.
answered Dec 2 '18 at 10:16
HermioneHermione
19619
$endgroup$
$begingroup$
Thanks for your answer thats exactly what I did however the equations got very messy. I was thinking if there is a more mathematically easier way of getting this solved
$endgroup$
– Hasan Hammoud
Dec 2 '18 at 10:21
add a comment |
$begingroup$
HINT: Let $A= (x_a , 0)$ and $B=(x_b, 0)$, with $x_a < x_b < 0$. You need to find $x_a$ and $x_b$. Consider a system of two equations:
In the first you write down the condition regarding the lenght ratio.
The second equation can be written using the Law of cosine, (see https://en.wikipedia.org/wiki/Law_of_cosines), since you know the cosine of the angle $ angle APB$.
answered Dec 2 '18 at 10:16
HermioneHermione
19619
$endgroup$
HINT: Let $A= (x_a , 0)$ and $B=(x_b, 0)$, with $x_a < x_b < 0$. You need to find $x_a$ and $x_b$. Consider a system of two equations:
In the first you write down the condition regarding the lenght ratio.
The second equation can be written using the Law of cosine, (see https://en.wikipedia.org/wiki/Law_of_cosines), since you know the cosine of the angle $ angle APB$.
answered Dec 2 '18 at 10:16
HermioneHermione
19619
answered Dec 2 '18 at 10:16
HermioneHermione
19619
answered Dec 2 '18 at 10:16
HermioneHermione
19619
answered Dec 2 '18 at 10:16
HermioneHermione
19619
19619
$begingroup$
Thanks for your answer thats exactly what I did however the equations got very messy. I was thinking if there is a more mathematically easier way of getting this solved
$endgroup$
– Hasan Hammoud
Dec 2 '18 at 10:21
add a comment |
$begingroup$
Thanks for your answer thats exactly what I did however the equations got very messy. I was thinking if there is a more mathematically easier way of getting this solved
$endgroup$
– Hasan Hammoud
Dec 2 '18 at 10:21
$begingroup$
Thanks for your answer thats exactly what I did however the equations got very messy. I was thinking if there is a more mathematically easier way of getting this solved
$endgroup$
– Hasan Hammoud
Dec 2 '18 at 10:21
$begingroup$
Thanks for your answer thats exactly what I did however the equations got very messy. I was thinking if there is a more mathematically easier way of getting this solved
$endgroup$
– Hasan Hammoud
Dec 2 '18 at 10:21
add a comment |
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1
$begingroup$
Isn't there any other condition or information? The problem as written has infinity solutions (A,B)
$endgroup$
– user376343
Dec 2 '18 at 9:45
1
$begingroup$
That information alone is not enough. You can rotate the triangle around point $P$ however you like and get infinitely many different values for $A$ and $B$. Or are you looking for a general expression?
$endgroup$
– platty
Dec 2 '18 at 9:46
$begingroup$
Sorry my bad you both are right I forgot to mention that A and B must be on the x-axis (-ve part of x-axis to be specific)
$endgroup$
– Hasan Hammoud
Dec 2 '18 at 9:48