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Point in right triangle given two points and an angle
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Given $triangle ABC$ with $angle B = 90^circ$, $overlineAC$ hypotenuse, known points $A = (x_a, y_a)$ and $B = (x_b, y_b)$, and known angle $angle A = theta$, how do I find $(x_c, y_c)$?
geometry trigonometry triangle
edited Aug 6 at 19:34
David G. Stork
7,7012929
asked Aug 6 at 19:32
Benn
1084
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Given $triangle ABC$ with $angle B = 90^circ$, $overlineAC$ hypotenuse, known points $A = (x_a, y_a)$ and $B = (x_b, y_b)$, and known angle $angle A = theta$, how do I find $(x_c, y_c)$?
geometry trigonometry triangle
edited Aug 6 at 19:34
David G. Stork
7,7012929
asked Aug 6 at 19:32
Benn
1084
You have not given enough information: How is $(x_c,y_c)$ defined, let alone $A$ and $B$?
– David G. Stork
Aug 6 at 19:36
I'm not sure I know what you mean. $(x_c, y_c)$ are the x- and y-coordinates of the point I'm trying to find, $A$ is an angle known ahead of time, and $B$ would then be $90 - A$ since this is a right triangle.
– Benn
Aug 6 at 19:39
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given $triangle ABC$ with $angle B = 90^circ$, $overlineAC$ hypotenuse, known points $A = (x_a, y_a)$ and $B = (x_b, y_b)$, and known angle $angle A = theta$, how do I find $(x_c, y_c)$?
geometry trigonometry triangle
edited Aug 6 at 19:34
David G. Stork
7,7012929
asked Aug 6 at 19:32
Benn
1084
Given $triangle ABC$ with $angle B = 90^circ$, $overlineAC$ hypotenuse, known points $A = (x_a, y_a)$ and $B = (x_b, y_b)$, and known angle $angle A = theta$, how do I find $(x_c, y_c)$?
geometry trigonometry triangle
edited Aug 6 at 19:34
David G. Stork
7,7012929
asked Aug 6 at 19:32
Benn
1084
edited Aug 6 at 19:34
David G. Stork
7,7012929
edited Aug 6 at 19:34
David G. Stork
7,7012929
edited Aug 6 at 19:34
David G. Stork
7,7012929
7,7012929
asked Aug 6 at 19:32
Benn
1084
asked Aug 6 at 19:32
Benn
1084
asked Aug 6 at 19:32
Benn
1084
1084
You have not given enough information: How is $(x_c,y_c)$ defined, let alone $A$ and $B$?
– David G. Stork
Aug 6 at 19:36
I'm not sure I know what you mean. $(x_c, y_c)$ are the x- and y-coordinates of the point I'm trying to find, $A$ is an angle known ahead of time, and $B$ would then be $90 - A$ since this is a right triangle.
– Benn
Aug 6 at 19:39
add a comment |Â
You have not given enough information: How is $(x_c,y_c)$ defined, let alone $A$ and $B$?
– David G. Stork
Aug 6 at 19:36
I'm not sure I know what you mean. $(x_c, y_c)$ are the x- and y-coordinates of the point I'm trying to find, $A$ is an angle known ahead of time, and $B$ would then be $90 - A$ since this is a right triangle.
– Benn
Aug 6 at 19:39
You have not given enough information: How is $(x_c,y_c)$ defined, let alone $A$ and $B$?
– David G. Stork
Aug 6 at 19:36
You have not given enough information: How is $(x_c,y_c)$ defined, let alone $A$ and $B$?
– David G. Stork
Aug 6 at 19:36
I'm not sure I know what you mean. $(x_c, y_c)$ are the x- and y-coordinates of the point I'm trying to find, $A$ is an angle known ahead of time, and $B$ would then be $90 - A$ since this is a right triangle.
– Benn
Aug 6 at 19:39
I'm not sure I know what you mean. $(x_c, y_c)$ are the x- and y-coordinates of the point I'm trying to find, $A$ is an angle known ahead of time, and $B$ would then be $90 - A$ since this is a right triangle.
– Benn
Aug 6 at 19:39
add a comment |Â
4 Answers
4
active
oldest
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up vote
2
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accepted
This is easiest with vectors. First calculate the vector from $B$ to $A$. This is simply:
$$v_BA=(x_a-x_b, y_a-y_b)$$
Rotate that vector by 90 degrees. You do this by swapping its coordinates and negating one of them. Which one you negate doesn't matter, and the choice determines which of the two solutions you get:
$$v^perp_BA = (-(y_a-y_b), x_a-x_b) text or (y_a-y_b, -(x_a-x_b))$$
Note however that the vector you now have is still of length $|BA|$, though it is now pointing in the right direction to $C$ (if you start from $B$).
To scale the vector to the right length we need to multiply it by the factor $fracBC = tantheta$:
$$v_BC = v^perp_BA cdot tantheta$$
Lastly you add this vector to the coordinates of $B$ to get the coordinates of $C$:
$$(x_c,y_c) = (x_b-(y_a-y_b)tantheta, y_b+(x_a-x_b)tantheta)$$
or
$$(x_c,y_c) = (x_b+(y_a-y_b)tantheta, y_b-(x_a-x_b)tantheta)$$
answered Aug 7 at 9:59
Jaap Scherphuis
3,508516
This is perfect and easy to implement efficiently. Thanks!
– Benn
Aug 7 at 13:24
Good explanation. Drawing may help to understand it better.
– yW0K5o
Aug 7 at 13:48
add a comment |Â
up vote
2
down vote
Write $$sin^2theta=(fracBCAC)^2=frac(x_c-x_b)^2+(y_c-y_b)^2(x_c-x_a)^2+(y_c-y_a)^2\cos^2theta=(fracABAC)^2=frac(x_b-x_a)^2+(y_b-y_a)^2(x_c-x_a)^2+(y_c-y_a)^2$$
You have two equations, with two unknowns. Notice that they are quadratic equations, so you will have two solutions, depending on which side of the $AB$ line you find point $C$.
edited Aug 7 at 9:12
yW0K5o
212211
answered Aug 6 at 19:42
Andrei
7,5852822
add a comment |Â
up vote
1
down vote
C is on the perpendicular to segment AB by B. So $x_c, y_c$ can be parametrized by $lambda$ : $x_c - x_b, y_c - y_b = lambda * (y_b - y_a, x_a - x_b)$
Now compute the angle of angle CAB as a function of $lambda$ and solve the equation (in $lambda$) angle= $theta$
answered Aug 6 at 19:42
Thomas
779612
Could you say more about this? I'm drawn to this approach as I'm going to end up implementing this as part of a drawing algorithm. $angle CAB$ will be defined as a constant up front (probably something like $10^circ$ or $15^circ$).
– Benn
Aug 7 at 0:56
from there you can use @Andrei's equation. Just replace $x_c$ and $y_c$ by the expression with lambda this way it makes a one unknown equation
– Thomas
Aug 7 at 2:12
add a comment |Â
up vote
0
down vote
Picture 1
Let's move system of axises to point $O'(x_a,y_b)$, the coordinates of the points $A, B, C$ will change respectively.
Picture 2
Le's name $angle O'AB=alpha$. Also $angle ABO' =90^o-alpha$ and sum of angles on axis X must be $180^o$, $angle CBX_c=alpha$.
From $triangle O'AB$ will find that $AB = fracy_b-y_acosalpha (1)$
From $triangle CBX_c$ will find that $BC = fracx_c-x_b+x_acosalpha (2)$
From $triangle ABC$ will find that $tgtheta= fracBCAB (3)$
Put (1) and (2) into (3)
$tgtheta= fracBCAB=fracfracx_c-x_b+x_acosalphafracy_b-y_acosalpha = fracx_c-x_b+x_ay_b-y_a(4)$
From (4) $x_c = (x_b-x_a) + (y_b-y_a)*tgtheta (5)$
If we move to the original axises (see Picture 1) we need to add $x_a$ to (5)
So (5) become $x_c = (x_b-x_a) + (y_b-y_a)*tgtheta + x_a= x_b + (y_b-y_a)*tgtheta (6)$
Let's find $y_c$ from the condition that $triangle O'AB$ is similar to $triangle X_cBC$ because of $angle O'AB = angle X_cBC = alpha$ and $angle AO'B = angle BX_cC = 90^o$
So $fracO'ABX_c=fracO'BX_cC (7)$ or
$fracy_b-y_ax_c-x_b+x_a=fracx_b-x_ay_c (7')$
From (7') $y_c=fracx_b-x_ay_b-y_a* (x_c-x_b+x_a) (7'')$
Put (5) to (7'')
$y_c=fracx_b-x_ay_b-y_a* ((x_b-x_a) + (y_b-y_a)*tgtheta-x_b+x_a) = (x_b-x_a)*tgtheta (7''')$
If we move to the original axises (see Picture 1) we need to add $y_b$ to (7''')
$y_c=(x_b-x_a)*tgtheta + y_b=y_b + (x_b-x_a)*tgtheta (8)$
Finding the second solution for $C'$ I leave to readers.
Questions, edit, comments?
answered Aug 8 at 1:52
yW0K5o
212211
Why do you align the triangle with the axes? The given points A and B can be anywhere and don't always have the same x-coordinate
– Jaap Scherphuis
Aug 8 at 5:27
Thank you for noticing this. I post the correct picture later.
– yW0K5o
Aug 8 at 7:54
I post new solution later.
– yW0K5o
Aug 8 at 23:08
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
This is easiest with vectors. First calculate the vector from $B$ to $A$. This is simply:
$$v_BA=(x_a-x_b, y_a-y_b)$$
Rotate that vector by 90 degrees. You do this by swapping its coordinates and negating one of them. Which one you negate doesn't matter, and the choice determines which of the two solutions you get:
$$v^perp_BA = (-(y_a-y_b), x_a-x_b) text or (y_a-y_b, -(x_a-x_b))$$
Note however that the vector you now have is still of length $|BA|$, though it is now pointing in the right direction to $C$ (if you start from $B$).
To scale the vector to the right length we need to multiply it by the factor $fracBC = tantheta$:
$$v_BC = v^perp_BA cdot tantheta$$
Lastly you add this vector to the coordinates of $B$ to get the coordinates of $C$:
$$(x_c,y_c) = (x_b-(y_a-y_b)tantheta, y_b+(x_a-x_b)tantheta)$$
or
$$(x_c,y_c) = (x_b+(y_a-y_b)tantheta, y_b-(x_a-x_b)tantheta)$$
answered Aug 7 at 9:59
Jaap Scherphuis
3,508516
This is perfect and easy to implement efficiently. Thanks!
– Benn
Aug 7 at 13:24
Good explanation. Drawing may help to understand it better.
– yW0K5o
Aug 7 at 13:48
add a comment |Â
up vote
2
down vote
accepted
This is easiest with vectors. First calculate the vector from $B$ to $A$. This is simply:
$$v_BA=(x_a-x_b, y_a-y_b)$$
Rotate that vector by 90 degrees. You do this by swapping its coordinates and negating one of them. Which one you negate doesn't matter, and the choice determines which of the two solutions you get:
$$v^perp_BA = (-(y_a-y_b), x_a-x_b) text or (y_a-y_b, -(x_a-x_b))$$
Note however that the vector you now have is still of length $|BA|$, though it is now pointing in the right direction to $C$ (if you start from $B$).
To scale the vector to the right length we need to multiply it by the factor $fracBC = tantheta$:
$$v_BC = v^perp_BA cdot tantheta$$
Lastly you add this vector to the coordinates of $B$ to get the coordinates of $C$:
$$(x_c,y_c) = (x_b-(y_a-y_b)tantheta, y_b+(x_a-x_b)tantheta)$$
or
$$(x_c,y_c) = (x_b+(y_a-y_b)tantheta, y_b-(x_a-x_b)tantheta)$$
answered Aug 7 at 9:59
Jaap Scherphuis
3,508516
This is perfect and easy to implement efficiently. Thanks!
– Benn
Aug 7 at 13:24
Good explanation. Drawing may help to understand it better.
– yW0K5o
Aug 7 at 13:48
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
This is easiest with vectors. First calculate the vector from $B$ to $A$. This is simply:
$$v_BA=(x_a-x_b, y_a-y_b)$$
Rotate that vector by 90 degrees. You do this by swapping its coordinates and negating one of them. Which one you negate doesn't matter, and the choice determines which of the two solutions you get:
$$v^perp_BA = (-(y_a-y_b), x_a-x_b) text or (y_a-y_b, -(x_a-x_b))$$
Note however that the vector you now have is still of length $|BA|$, though it is now pointing in the right direction to $C$ (if you start from $B$).
To scale the vector to the right length we need to multiply it by the factor $fracBC = tantheta$:
$$v_BC = v^perp_BA cdot tantheta$$
Lastly you add this vector to the coordinates of $B$ to get the coordinates of $C$:
$$(x_c,y_c) = (x_b-(y_a-y_b)tantheta, y_b+(x_a-x_b)tantheta)$$
or
$$(x_c,y_c) = (x_b+(y_a-y_b)tantheta, y_b-(x_a-x_b)tantheta)$$
answered Aug 7 at 9:59
Jaap Scherphuis
3,508516
This is easiest with vectors. First calculate the vector from $B$ to $A$. This is simply:
$$v_BA=(x_a-x_b, y_a-y_b)$$
Rotate that vector by 90 degrees. You do this by swapping its coordinates and negating one of them. Which one you negate doesn't matter, and the choice determines which of the two solutions you get:
$$v^perp_BA = (-(y_a-y_b), x_a-x_b) text or (y_a-y_b, -(x_a-x_b))$$
Note however that the vector you now have is still of length $|BA|$, though it is now pointing in the right direction to $C$ (if you start from $B$).
To scale the vector to the right length we need to multiply it by the factor $fracBC = tantheta$:
$$v_BC = v^perp_BA cdot tantheta$$
Lastly you add this vector to the coordinates of $B$ to get the coordinates of $C$:
$$(x_c,y_c) = (x_b-(y_a-y_b)tantheta, y_b+(x_a-x_b)tantheta)$$
or
$$(x_c,y_c) = (x_b+(y_a-y_b)tantheta, y_b-(x_a-x_b)tantheta)$$
answered Aug 7 at 9:59
Jaap Scherphuis
3,508516
edited Aug 7 at 10:05
answered Aug 7 at 9:59
Jaap Scherphuis
3,508516
answered Aug 7 at 9:59
Jaap Scherphuis
3,508516
answered Aug 7 at 9:59
Jaap Scherphuis
3,508516
3,508516
This is perfect and easy to implement efficiently. Thanks!
– Benn
Aug 7 at 13:24
Good explanation. Drawing may help to understand it better.
– yW0K5o
Aug 7 at 13:48
add a comment |Â
This is perfect and easy to implement efficiently. Thanks!
– Benn
Aug 7 at 13:24
Good explanation. Drawing may help to understand it better.
– yW0K5o
Aug 7 at 13:48
This is perfect and easy to implement efficiently. Thanks!
– Benn
Aug 7 at 13:24
This is perfect and easy to implement efficiently. Thanks!
– Benn
Aug 7 at 13:24
Good explanation. Drawing may help to understand it better.
– yW0K5o
Aug 7 at 13:48
Good explanation. Drawing may help to understand it better.
– yW0K5o
Aug 7 at 13:48
add a comment |Â
up vote
2
down vote
Write $$sin^2theta=(fracBCAC)^2=frac(x_c-x_b)^2+(y_c-y_b)^2(x_c-x_a)^2+(y_c-y_a)^2\cos^2theta=(fracABAC)^2=frac(x_b-x_a)^2+(y_b-y_a)^2(x_c-x_a)^2+(y_c-y_a)^2$$
You have two equations, with two unknowns. Notice that they are quadratic equations, so you will have two solutions, depending on which side of the $AB$ line you find point $C$.
edited Aug 7 at 9:12
yW0K5o
212211
answered Aug 6 at 19:42
Andrei
7,5852822
add a comment |Â
up vote
2
down vote
Write $$sin^2theta=(fracBCAC)^2=frac(x_c-x_b)^2+(y_c-y_b)^2(x_c-x_a)^2+(y_c-y_a)^2\cos^2theta=(fracABAC)^2=frac(x_b-x_a)^2+(y_b-y_a)^2(x_c-x_a)^2+(y_c-y_a)^2$$
You have two equations, with two unknowns. Notice that they are quadratic equations, so you will have two solutions, depending on which side of the $AB$ line you find point $C$.
edited Aug 7 at 9:12
yW0K5o
212211
answered Aug 6 at 19:42
Andrei
7,5852822
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Write $$sin^2theta=(fracBCAC)^2=frac(x_c-x_b)^2+(y_c-y_b)^2(x_c-x_a)^2+(y_c-y_a)^2\cos^2theta=(fracABAC)^2=frac(x_b-x_a)^2+(y_b-y_a)^2(x_c-x_a)^2+(y_c-y_a)^2$$
You have two equations, with two unknowns. Notice that they are quadratic equations, so you will have two solutions, depending on which side of the $AB$ line you find point $C$.
edited Aug 7 at 9:12
yW0K5o
212211
answered Aug 6 at 19:42
Andrei
7,5852822
Write $$sin^2theta=(fracBCAC)^2=frac(x_c-x_b)^2+(y_c-y_b)^2(x_c-x_a)^2+(y_c-y_a)^2\cos^2theta=(fracABAC)^2=frac(x_b-x_a)^2+(y_b-y_a)^2(x_c-x_a)^2+(y_c-y_a)^2$$
You have two equations, with two unknowns. Notice that they are quadratic equations, so you will have two solutions, depending on which side of the $AB$ line you find point $C$.
edited Aug 7 at 9:12
yW0K5o
212211
answered Aug 6 at 19:42
Andrei
7,5852822
edited Aug 7 at 9:12
yW0K5o
212211
edited Aug 7 at 9:12
yW0K5o
212211
edited Aug 7 at 9:12
yW0K5o
212211
212211
answered Aug 6 at 19:42
Andrei
7,5852822
answered Aug 6 at 19:42
Andrei
7,5852822
answered Aug 6 at 19:42
Andrei
7,5852822
7,5852822
add a comment |Â
add a comment |Â
up vote
1
down vote
C is on the perpendicular to segment AB by B. So $x_c, y_c$ can be parametrized by $lambda$ : $x_c - x_b, y_c - y_b = lambda * (y_b - y_a, x_a - x_b)$
Now compute the angle of angle CAB as a function of $lambda$ and solve the equation (in $lambda$) angle= $theta$
answered Aug 6 at 19:42
Thomas
779612
Could you say more about this? I'm drawn to this approach as I'm going to end up implementing this as part of a drawing algorithm. $angle CAB$ will be defined as a constant up front (probably something like $10^circ$ or $15^circ$).
– Benn
Aug 7 at 0:56
from there you can use @Andrei's equation. Just replace $x_c$ and $y_c$ by the expression with lambda this way it makes a one unknown equation
– Thomas
Aug 7 at 2:12
add a comment |Â
up vote
1
down vote
C is on the perpendicular to segment AB by B. So $x_c, y_c$ can be parametrized by $lambda$ : $x_c - x_b, y_c - y_b = lambda * (y_b - y_a, x_a - x_b)$
Now compute the angle of angle CAB as a function of $lambda$ and solve the equation (in $lambda$) angle= $theta$
answered Aug 6 at 19:42
Thomas
779612
Could you say more about this? I'm drawn to this approach as I'm going to end up implementing this as part of a drawing algorithm. $angle CAB$ will be defined as a constant up front (probably something like $10^circ$ or $15^circ$).
– Benn
Aug 7 at 0:56
from there you can use @Andrei's equation. Just replace $x_c$ and $y_c$ by the expression with lambda this way it makes a one unknown equation
– Thomas
Aug 7 at 2:12
add a comment |Â
up vote
1
down vote
up vote
1
down vote
C is on the perpendicular to segment AB by B. So $x_c, y_c$ can be parametrized by $lambda$ : $x_c - x_b, y_c - y_b = lambda * (y_b - y_a, x_a - x_b)$
Now compute the angle of angle CAB as a function of $lambda$ and solve the equation (in $lambda$) angle= $theta$
answered Aug 6 at 19:42
Thomas
779612
C is on the perpendicular to segment AB by B. So $x_c, y_c$ can be parametrized by $lambda$ : $x_c - x_b, y_c - y_b = lambda * (y_b - y_a, x_a - x_b)$
Now compute the angle of angle CAB as a function of $lambda$ and solve the equation (in $lambda$) angle= $theta$
answered Aug 6 at 19:42
Thomas
779612
answered Aug 6 at 19:42
Thomas
779612
answered Aug 6 at 19:42
Thomas
779612
answered Aug 6 at 19:42
Thomas
779612
779612
Could you say more about this? I'm drawn to this approach as I'm going to end up implementing this as part of a drawing algorithm. $angle CAB$ will be defined as a constant up front (probably something like $10^circ$ or $15^circ$).
– Benn
Aug 7 at 0:56
from there you can use @Andrei's equation. Just replace $x_c$ and $y_c$ by the expression with lambda this way it makes a one unknown equation
– Thomas
Aug 7 at 2:12
add a comment |Â
Could you say more about this? I'm drawn to this approach as I'm going to end up implementing this as part of a drawing algorithm. $angle CAB$ will be defined as a constant up front (probably something like $10^circ$ or $15^circ$).
– Benn
Aug 7 at 0:56
from there you can use @Andrei's equation. Just replace $x_c$ and $y_c$ by the expression with lambda this way it makes a one unknown equation
– Thomas
Aug 7 at 2:12
Could you say more about this? I'm drawn to this approach as I'm going to end up implementing this as part of a drawing algorithm. $angle CAB$ will be defined as a constant up front (probably something like $10^circ$ or $15^circ$).
– Benn
Aug 7 at 0:56
Could you say more about this? I'm drawn to this approach as I'm going to end up implementing this as part of a drawing algorithm. $angle CAB$ will be defined as a constant up front (probably something like $10^circ$ or $15^circ$).
– Benn
Aug 7 at 0:56
from there you can use @Andrei's equation. Just replace $x_c$ and $y_c$ by the expression with lambda this way it makes a one unknown equation
– Thomas
Aug 7 at 2:12
from there you can use @Andrei's equation. Just replace $x_c$ and $y_c$ by the expression with lambda this way it makes a one unknown equation
– Thomas
Aug 7 at 2:12
add a comment |Â
up vote
0
down vote
Picture 1
Let's move system of axises to point $O'(x_a,y_b)$, the coordinates of the points $A, B, C$ will change respectively.
Picture 2
Le's name $angle O'AB=alpha$. Also $angle ABO' =90^o-alpha$ and sum of angles on axis X must be $180^o$, $angle CBX_c=alpha$.
From $triangle O'AB$ will find that $AB = fracy_b-y_acosalpha (1)$
From $triangle CBX_c$ will find that $BC = fracx_c-x_b+x_acosalpha (2)$
From $triangle ABC$ will find that $tgtheta= fracBCAB (3)$
Put (1) and (2) into (3)
$tgtheta= fracBCAB=fracfracx_c-x_b+x_acosalphafracy_b-y_acosalpha = fracx_c-x_b+x_ay_b-y_a(4)$
From (4) $x_c = (x_b-x_a) + (y_b-y_a)*tgtheta (5)$
If we move to the original axises (see Picture 1) we need to add $x_a$ to (5)
So (5) become $x_c = (x_b-x_a) + (y_b-y_a)*tgtheta + x_a= x_b + (y_b-y_a)*tgtheta (6)$
Let's find $y_c$ from the condition that $triangle O'AB$ is similar to $triangle X_cBC$ because of $angle O'AB = angle X_cBC = alpha$ and $angle AO'B = angle BX_cC = 90^o$
So $fracO'ABX_c=fracO'BX_cC (7)$ or
$fracy_b-y_ax_c-x_b+x_a=fracx_b-x_ay_c (7')$
From (7') $y_c=fracx_b-x_ay_b-y_a* (x_c-x_b+x_a) (7'')$
Put (5) to (7'')
$y_c=fracx_b-x_ay_b-y_a* ((x_b-x_a) + (y_b-y_a)*tgtheta-x_b+x_a) = (x_b-x_a)*tgtheta (7''')$
If we move to the original axises (see Picture 1) we need to add $y_b$ to (7''')
$y_c=(x_b-x_a)*tgtheta + y_b=y_b + (x_b-x_a)*tgtheta (8)$
Finding the second solution for $C'$ I leave to readers.
Questions, edit, comments?
answered Aug 8 at 1:52
yW0K5o
212211
Why do you align the triangle with the axes? The given points A and B can be anywhere and don't always have the same x-coordinate
– Jaap Scherphuis
Aug 8 at 5:27
Thank you for noticing this. I post the correct picture later.
– yW0K5o
Aug 8 at 7:54
I post new solution later.
– yW0K5o
Aug 8 at 23:08
add a comment |Â
up vote
0
down vote
Picture 1
Let's move system of axises to point $O'(x_a,y_b)$, the coordinates of the points $A, B, C$ will change respectively.
Picture 2
Le's name $angle O'AB=alpha$. Also $angle ABO' =90^o-alpha$ and sum of angles on axis X must be $180^o$, $angle CBX_c=alpha$.
From $triangle O'AB$ will find that $AB = fracy_b-y_acosalpha (1)$
From $triangle CBX_c$ will find that $BC = fracx_c-x_b+x_acosalpha (2)$
From $triangle ABC$ will find that $tgtheta= fracBCAB (3)$
Put (1) and (2) into (3)
$tgtheta= fracBCAB=fracfracx_c-x_b+x_acosalphafracy_b-y_acosalpha = fracx_c-x_b+x_ay_b-y_a(4)$
From (4) $x_c = (x_b-x_a) + (y_b-y_a)*tgtheta (5)$
If we move to the original axises (see Picture 1) we need to add $x_a$ to (5)
So (5) become $x_c = (x_b-x_a) + (y_b-y_a)*tgtheta + x_a= x_b + (y_b-y_a)*tgtheta (6)$
Let's find $y_c$ from the condition that $triangle O'AB$ is similar to $triangle X_cBC$ because of $angle O'AB = angle X_cBC = alpha$ and $angle AO'B = angle BX_cC = 90^o$
So $fracO'ABX_c=fracO'BX_cC (7)$ or
$fracy_b-y_ax_c-x_b+x_a=fracx_b-x_ay_c (7')$
From (7') $y_c=fracx_b-x_ay_b-y_a* (x_c-x_b+x_a) (7'')$
Put (5) to (7'')
$y_c=fracx_b-x_ay_b-y_a* ((x_b-x_a) + (y_b-y_a)*tgtheta-x_b+x_a) = (x_b-x_a)*tgtheta (7''')$
If we move to the original axises (see Picture 1) we need to add $y_b$ to (7''')
$y_c=(x_b-x_a)*tgtheta + y_b=y_b + (x_b-x_a)*tgtheta (8)$
Finding the second solution for $C'$ I leave to readers.
Questions, edit, comments?
answered Aug 8 at 1:52
yW0K5o
212211
Why do you align the triangle with the axes? The given points A and B can be anywhere and don't always have the same x-coordinate
– Jaap Scherphuis
Aug 8 at 5:27
Thank you for noticing this. I post the correct picture later.
– yW0K5o
Aug 8 at 7:54
I post new solution later.
– yW0K5o
Aug 8 at 23:08
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Picture 1
Let's move system of axises to point $O'(x_a,y_b)$, the coordinates of the points $A, B, C$ will change respectively.
Picture 2
Le's name $angle O'AB=alpha$. Also $angle ABO' =90^o-alpha$ and sum of angles on axis X must be $180^o$, $angle CBX_c=alpha$.
From $triangle O'AB$ will find that $AB = fracy_b-y_acosalpha (1)$
From $triangle CBX_c$ will find that $BC = fracx_c-x_b+x_acosalpha (2)$
From $triangle ABC$ will find that $tgtheta= fracBCAB (3)$
Put (1) and (2) into (3)
$tgtheta= fracBCAB=fracfracx_c-x_b+x_acosalphafracy_b-y_acosalpha = fracx_c-x_b+x_ay_b-y_a(4)$
From (4) $x_c = (x_b-x_a) + (y_b-y_a)*tgtheta (5)$
If we move to the original axises (see Picture 1) we need to add $x_a$ to (5)
So (5) become $x_c = (x_b-x_a) + (y_b-y_a)*tgtheta + x_a= x_b + (y_b-y_a)*tgtheta (6)$
Let's find $y_c$ from the condition that $triangle O'AB$ is similar to $triangle X_cBC$ because of $angle O'AB = angle X_cBC = alpha$ and $angle AO'B = angle BX_cC = 90^o$
So $fracO'ABX_c=fracO'BX_cC (7)$ or
$fracy_b-y_ax_c-x_b+x_a=fracx_b-x_ay_c (7')$
From (7') $y_c=fracx_b-x_ay_b-y_a* (x_c-x_b+x_a) (7'')$
Put (5) to (7'')
$y_c=fracx_b-x_ay_b-y_a* ((x_b-x_a) + (y_b-y_a)*tgtheta-x_b+x_a) = (x_b-x_a)*tgtheta (7''')$
If we move to the original axises (see Picture 1) we need to add $y_b$ to (7''')
$y_c=(x_b-x_a)*tgtheta + y_b=y_b + (x_b-x_a)*tgtheta (8)$
Finding the second solution for $C'$ I leave to readers.
Questions, edit, comments?
answered Aug 8 at 1:52
yW0K5o
212211
Picture 1
Let's move system of axises to point $O'(x_a,y_b)$, the coordinates of the points $A, B, C$ will change respectively.
Picture 2
Le's name $angle O'AB=alpha$. Also $angle ABO' =90^o-alpha$ and sum of angles on axis X must be $180^o$, $angle CBX_c=alpha$.
From $triangle O'AB$ will find that $AB = fracy_b-y_acosalpha (1)$
From $triangle CBX_c$ will find that $BC = fracx_c-x_b+x_acosalpha (2)$
From $triangle ABC$ will find that $tgtheta= fracBCAB (3)$
Put (1) and (2) into (3)
$tgtheta= fracBCAB=fracfracx_c-x_b+x_acosalphafracy_b-y_acosalpha = fracx_c-x_b+x_ay_b-y_a(4)$
From (4) $x_c = (x_b-x_a) + (y_b-y_a)*tgtheta (5)$
If we move to the original axises (see Picture 1) we need to add $x_a$ to (5)
So (5) become $x_c = (x_b-x_a) + (y_b-y_a)*tgtheta + x_a= x_b + (y_b-y_a)*tgtheta (6)$
Let's find $y_c$ from the condition that $triangle O'AB$ is similar to $triangle X_cBC$ because of $angle O'AB = angle X_cBC = alpha$ and $angle AO'B = angle BX_cC = 90^o$
So $fracO'ABX_c=fracO'BX_cC (7)$ or
$fracy_b-y_ax_c-x_b+x_a=fracx_b-x_ay_c (7')$
From (7') $y_c=fracx_b-x_ay_b-y_a* (x_c-x_b+x_a) (7'')$
Put (5) to (7'')
$y_c=fracx_b-x_ay_b-y_a* ((x_b-x_a) + (y_b-y_a)*tgtheta-x_b+x_a) = (x_b-x_a)*tgtheta (7''')$
If we move to the original axises (see Picture 1) we need to add $y_b$ to (7''')
$y_c=(x_b-x_a)*tgtheta + y_b=y_b + (x_b-x_a)*tgtheta (8)$
Finding the second solution for $C'$ I leave to readers.
Questions, edit, comments?
answered Aug 8 at 1:52
yW0K5o
212211
edited Aug 9 at 0:35
answered Aug 8 at 1:52
yW0K5o
212211
answered Aug 8 at 1:52
yW0K5o
212211
answered Aug 8 at 1:52
yW0K5o
212211
212211
Why do you align the triangle with the axes? The given points A and B can be anywhere and don't always have the same x-coordinate
– Jaap Scherphuis
Aug 8 at 5:27
Thank you for noticing this. I post the correct picture later.
– yW0K5o
Aug 8 at 7:54
I post new solution later.
– yW0K5o
Aug 8 at 23:08
add a comment |Â
Why do you align the triangle with the axes? The given points A and B can be anywhere and don't always have the same x-coordinate
– Jaap Scherphuis
Aug 8 at 5:27
Thank you for noticing this. I post the correct picture later.
– yW0K5o
Aug 8 at 7:54
I post new solution later.
– yW0K5o
Aug 8 at 23:08
Why do you align the triangle with the axes? The given points A and B can be anywhere and don't always have the same x-coordinate
– Jaap Scherphuis
Aug 8 at 5:27
Why do you align the triangle with the axes? The given points A and B can be anywhere and don't always have the same x-coordinate
– Jaap Scherphuis
Aug 8 at 5:27
Thank you for noticing this. I post the correct picture later.
– yW0K5o
Aug 8 at 7:54
Thank you for noticing this. I post the correct picture later.
– yW0K5o
Aug 8 at 7:54
I post new solution later.
– yW0K5o
Aug 8 at 23:08
I post new solution later.
– yW0K5o
Aug 8 at 23:08
add a comment |Â
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You have not given enough information: How is $(x_c,y_c)$ defined, let alone $A$ and $B$?
– David G. Stork
Aug 6 at 19:36
I'm not sure I know what you mean. $(x_c, y_c)$ are the x- and y-coordinates of the point I'm trying to find, $A$ is an angle known ahead of time, and $B$ would then be $90 - A$ since this is a right triangle.
– Benn
Aug 6 at 19:39