2D Coordinate - Given two lines and their angles, compute the point in which the perpendicular foot length...
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I am given values of $x_1, y_1, x_2, y_2$, $theta_{1}$, $theta_{2}$, and $h$.
I want to find out the coordinate of the Unknown Point of Interest.
The Unknown Point of Interest is the point in which the length of the perpendicular foot between the two blue lines with $theta_{1}$ becomes $h$. How can I do this?
geometry trigonometry
edited Nov 26 at 4:45
Boshu
700315
asked Nov 26 at 4:37
Eric Kim
1073
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up vote
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down vote
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I am given values of $x_1, y_1, x_2, y_2$, $theta_{1}$, $theta_{2}$, and $h$.
I want to find out the coordinate of the Unknown Point of Interest.
The Unknown Point of Interest is the point in which the length of the perpendicular foot between the two blue lines with $theta_{1}$ becomes $h$. How can I do this?
geometry trigonometry
edited Nov 26 at 4:45
Boshu
700315
asked Nov 26 at 4:37
Eric Kim
1073
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am given values of $x_1, y_1, x_2, y_2$, $theta_{1}$, $theta_{2}$, and $h$.
I want to find out the coordinate of the Unknown Point of Interest.
The Unknown Point of Interest is the point in which the length of the perpendicular foot between the two blue lines with $theta_{1}$ becomes $h$. How can I do this?
geometry trigonometry
edited Nov 26 at 4:45
Boshu
700315
asked Nov 26 at 4:37
Eric Kim
1073
I am given values of $x_1, y_1, x_2, y_2$, $theta_{1}$, $theta_{2}$, and $h$.
I want to find out the coordinate of the Unknown Point of Interest.
The Unknown Point of Interest is the point in which the length of the perpendicular foot between the two blue lines with $theta_{1}$ becomes $h$. How can I do this?
geometry trigonometry
geometry trigonometry
edited Nov 26 at 4:45
Boshu
700315
asked Nov 26 at 4:37
Eric Kim
1073
edited Nov 26 at 4:45
Boshu
700315
asked Nov 26 at 4:37
Eric Kim
1073
edited Nov 26 at 4:45
Boshu
700315
edited Nov 26 at 4:45
Boshu
700315
edited Nov 26 at 4:45
Boshu
700315
700315
asked Nov 26 at 4:37
Eric Kim
1073
asked Nov 26 at 4:37
Eric Kim
1073
asked Nov 26 at 4:37
Eric Kim
1073
1073
add a comment |
add a comment |
1 Answer
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Use trigonometry. If you know $h$ and the angle $theta_1$, then you can treat the triangle as a right angled triangle and the length of the hypotenuse as $frac{h}{sintheta_1}=h_1$. Then the point of interest can be written as ${h_1cos(theta_1+theta_2),h_1sin(theta_1+theta_2)}$ using the same trigonometric argument/transformation from polar co-ordinates to Cartesian.
answered Nov 26 at 4:42
Boshu
700315
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
up vote
2
down vote
accepted
Use trigonometry. If you know $h$ and the angle $theta_1$, then you can treat the triangle as a right angled triangle and the length of the hypotenuse as $frac{h}{sintheta_1}=h_1$. Then the point of interest can be written as ${h_1cos(theta_1+theta_2),h_1sin(theta_1+theta_2)}$ using the same trigonometric argument/transformation from polar co-ordinates to Cartesian.
answered Nov 26 at 4:42
Boshu
700315
add a comment |
up vote
2
down vote
accepted
Use trigonometry. If you know $h$ and the angle $theta_1$, then you can treat the triangle as a right angled triangle and the length of the hypotenuse as $frac{h}{sintheta_1}=h_1$. Then the point of interest can be written as ${h_1cos(theta_1+theta_2),h_1sin(theta_1+theta_2)}$ using the same trigonometric argument/transformation from polar co-ordinates to Cartesian.
answered Nov 26 at 4:42
Boshu
700315
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Use trigonometry. If you know $h$ and the angle $theta_1$, then you can treat the triangle as a right angled triangle and the length of the hypotenuse as $frac{h}{sintheta_1}=h_1$. Then the point of interest can be written as ${h_1cos(theta_1+theta_2),h_1sin(theta_1+theta_2)}$ using the same trigonometric argument/transformation from polar co-ordinates to Cartesian.
answered Nov 26 at 4:42
Boshu
700315
Use trigonometry. If you know $h$ and the angle $theta_1$, then you can treat the triangle as a right angled triangle and the length of the hypotenuse as $frac{h}{sintheta_1}=h_1$. Then the point of interest can be written as ${h_1cos(theta_1+theta_2),h_1sin(theta_1+theta_2)}$ using the same trigonometric argument/transformation from polar co-ordinates to Cartesian.
answered Nov 26 at 4:42
Boshu
700315
edited Nov 26 at 4:54
answered Nov 26 at 4:42
Boshu
700315
answered Nov 26 at 4:42
Boshu
700315
answered Nov 26 at 4:42
Boshu
700315
700315
add a comment |
add a comment |
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