Determine vertex coordinates of a triangle if length and angles of opposite are known
$begingroup$
Given a triangle such as this:
Where $C$, $A$ and $B$ are cartesian coordinates and $a$, $b$, $c$ are the lengths of the sides.
I know that $$C = (bcostheta,;bsintheta)$$
where $theta$ is the angle at vertex $A$.
However, in my case, I do not know $b$, but I do know $c$ and the gradient at both $A$ and $B$.
How can I find $C$?
trigonometry triangle
edited Jan 15 '16 at 13:51
JnxF
1,0691821
asked Jan 15 '16 at 13:36
jrammjramm
1014
$endgroup$
add a comment |
$begingroup$
Given a triangle such as this:
Where $C$, $A$ and $B$ are cartesian coordinates and $a$, $b$, $c$ are the lengths of the sides.
I know that $$C = (bcostheta,;bsintheta)$$
where $theta$ is the angle at vertex $A$.
However, in my case, I do not know $b$, but I do know $c$ and the gradient at both $A$ and $B$.
How can I find $C$?
trigonometry triangle
edited Jan 15 '16 at 13:51
JnxF
1,0691821
asked Jan 15 '16 at 13:36
jrammjramm
1014
$endgroup$
add a comment |
$begingroup$
Given a triangle such as this:
Where $C$, $A$ and $B$ are cartesian coordinates and $a$, $b$, $c$ are the lengths of the sides.
I know that $$C = (bcostheta,;bsintheta)$$
where $theta$ is the angle at vertex $A$.
However, in my case, I do not know $b$, but I do know $c$ and the gradient at both $A$ and $B$.
How can I find $C$?
trigonometry triangle
edited Jan 15 '16 at 13:51
JnxF
1,0691821
asked Jan 15 '16 at 13:36
jrammjramm
1014
$endgroup$
Given a triangle such as this:
Where $C$, $A$ and $B$ are cartesian coordinates and $a$, $b$, $c$ are the lengths of the sides.
I know that $$C = (bcostheta,;bsintheta)$$
where $theta$ is the angle at vertex $A$.
However, in my case, I do not know $b$, but I do know $c$ and the gradient at both $A$ and $B$.
How can I find $C$?
trigonometry triangle
trigonometry triangle
edited Jan 15 '16 at 13:51
JnxF
1,0691821
asked Jan 15 '16 at 13:36
jrammjramm
1014
edited Jan 15 '16 at 13:51
JnxF
1,0691821
asked Jan 15 '16 at 13:36
jrammjramm
1014
edited Jan 15 '16 at 13:51
JnxF
1,0691821
edited Jan 15 '16 at 13:51
JnxF
1,0691821
edited Jan 15 '16 at 13:51
JnxF
1,0691821
1,0691821
asked Jan 15 '16 at 13:36
jrammjramm
1014
asked Jan 15 '16 at 13:36
jrammjramm
1014
asked Jan 15 '16 at 13:36
jrammjramm
1014
1014
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The equation $C = (bcostheta,;bsintheta)$ makes sense only if
the coordinates of $A$ are $(0,0)$, so I suppose $A = (0,0)$.
Then $B = (c,0)$.
From a point and the slope of a line, you can derive an equation
for the line. So you can easily find equations for lines $AC$ and $CB$.
From the equations of two lines you can find the coordinates of
their intersection.
That should be enough to solve the problem. But if the slope of $AC$ is
$tan alpha$ and the slope of $CB$ is $-tan beta$
(where $alpha$ and $beta$ are the positive angles at $A$ and $B$,
and the slope of $CB$ is negative indicating it slopes down to the right),
then the coordinates of $C=(c_x, c_y)$ can be found from the
relations
begin{align}
c_y &= c_x tan alpha, \
c_y &= (c - c_x)tan beta,
end{align}
as you can confirm by dropping a perpendicular from $C$ to $AB$ and
labeling the lengths of the legs and angles $alpha$ and $beta$
of the two right triangles so drawn.
answered Jan 15 '16 at 15:26
David KDavid K
55.1k344120
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The equation $C = (bcostheta,;bsintheta)$ makes sense only if
the coordinates of $A$ are $(0,0)$, so I suppose $A = (0,0)$.
Then $B = (c,0)$.
From a point and the slope of a line, you can derive an equation
for the line. So you can easily find equations for lines $AC$ and $CB$.
From the equations of two lines you can find the coordinates of
their intersection.
That should be enough to solve the problem. But if the slope of $AC$ is
$tan alpha$ and the slope of $CB$ is $-tan beta$
(where $alpha$ and $beta$ are the positive angles at $A$ and $B$,
and the slope of $CB$ is negative indicating it slopes down to the right),
then the coordinates of $C=(c_x, c_y)$ can be found from the
relations
begin{align}
c_y &= c_x tan alpha, \
c_y &= (c - c_x)tan beta,
end{align}
as you can confirm by dropping a perpendicular from $C$ to $AB$ and
labeling the lengths of the legs and angles $alpha$ and $beta$
of the two right triangles so drawn.
answered Jan 15 '16 at 15:26
David KDavid K
55.1k344120
$endgroup$
add a comment |
$begingroup$
The equation $C = (bcostheta,;bsintheta)$ makes sense only if
the coordinates of $A$ are $(0,0)$, so I suppose $A = (0,0)$.
Then $B = (c,0)$.
From a point and the slope of a line, you can derive an equation
for the line. So you can easily find equations for lines $AC$ and $CB$.
From the equations of two lines you can find the coordinates of
their intersection.
That should be enough to solve the problem. But if the slope of $AC$ is
$tan alpha$ and the slope of $CB$ is $-tan beta$
(where $alpha$ and $beta$ are the positive angles at $A$ and $B$,
and the slope of $CB$ is negative indicating it slopes down to the right),
then the coordinates of $C=(c_x, c_y)$ can be found from the
relations
begin{align}
c_y &= c_x tan alpha, \
c_y &= (c - c_x)tan beta,
end{align}
as you can confirm by dropping a perpendicular from $C$ to $AB$ and
labeling the lengths of the legs and angles $alpha$ and $beta$
of the two right triangles so drawn.
answered Jan 15 '16 at 15:26
David KDavid K
55.1k344120
$endgroup$
add a comment |
$begingroup$
The equation $C = (bcostheta,;bsintheta)$ makes sense only if
the coordinates of $A$ are $(0,0)$, so I suppose $A = (0,0)$.
Then $B = (c,0)$.
From a point and the slope of a line, you can derive an equation
for the line. So you can easily find equations for lines $AC$ and $CB$.
From the equations of two lines you can find the coordinates of
their intersection.
That should be enough to solve the problem. But if the slope of $AC$ is
$tan alpha$ and the slope of $CB$ is $-tan beta$
(where $alpha$ and $beta$ are the positive angles at $A$ and $B$,
and the slope of $CB$ is negative indicating it slopes down to the right),
then the coordinates of $C=(c_x, c_y)$ can be found from the
relations
begin{align}
c_y &= c_x tan alpha, \
c_y &= (c - c_x)tan beta,
end{align}
as you can confirm by dropping a perpendicular from $C$ to $AB$ and
labeling the lengths of the legs and angles $alpha$ and $beta$
of the two right triangles so drawn.
answered Jan 15 '16 at 15:26
David KDavid K
55.1k344120
$endgroup$
The equation $C = (bcostheta,;bsintheta)$ makes sense only if
the coordinates of $A$ are $(0,0)$, so I suppose $A = (0,0)$.
Then $B = (c,0)$.
From a point and the slope of a line, you can derive an equation
for the line. So you can easily find equations for lines $AC$ and $CB$.
From the equations of two lines you can find the coordinates of
their intersection.
That should be enough to solve the problem. But if the slope of $AC$ is
$tan alpha$ and the slope of $CB$ is $-tan beta$
(where $alpha$ and $beta$ are the positive angles at $A$ and $B$,
and the slope of $CB$ is negative indicating it slopes down to the right),
then the coordinates of $C=(c_x, c_y)$ can be found from the
relations
begin{align}
c_y &= c_x tan alpha, \
c_y &= (c - c_x)tan beta,
end{align}
as you can confirm by dropping a perpendicular from $C$ to $AB$ and
labeling the lengths of the legs and angles $alpha$ and $beta$
of the two right triangles so drawn.
answered Jan 15 '16 at 15:26
David KDavid K
55.1k344120
answered Jan 15 '16 at 15:26
David KDavid K
55.1k344120
answered Jan 15 '16 at 15:26
David KDavid K
55.1k344120
answered Jan 15 '16 at 15:26
David KDavid K
55.1k344120
55.1k344120
add a comment |
add a comment |
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