Graphing: plot third point of triangle, given all sides and angles

[galfridus]

Hello all, I have a question about plotting triangles on a graph.

The problem:

Given all three side lengths and angles, and two points, find the position of the third point.

Parameters
Points:
A=[0,0]
B=[5,15]
C=?

Sides (length):
A-B (a) = sqrt( (5-0)^2 + (15-0)^2) = 15.81
A-C (b) = 5
B-C (c) = 14

Angles:
A = 102
B = 18
C = 60

Given all of this information, I know it's possible to plot the third point, but I can't seem to come up with a function to describe it.

Any suggestions would be greatly appreciated - I've been stuck on this for a few weeks now...
(I apologise for formatting issues, posting from a mobile device)

 

[DrPhil]

Hello all, I have a question about plotting triangles on a graph.

The problem:

Given all three side lengths and angles, and two points, find the position of the third point.

Parameters
Points:
A=[0,0]
B=[5,15]
C=?

Sides (length):
A-B (a) = sqrt( (5-0)^2 + (15-0)^2) = 15.81
A-C (b) = 5
B-C (c) = 14

Angles:
A = 102
B = 18
C = 60

Given all of this information, I know it's possible to plot the third point, but I can't seem to come up with a function to describe it.

Any suggestions would be greatly appreciated - I've been stuck on this for a few weeks now...
(I apologise for formatting issues, posting from a mobile device)

You have TOO MUCH information - and not everything agrees. In particular, are you really given that b=5? That doesn't seem to "fit." Please give us the problem as presented.

Just telling us that your don't have the right answer doesn't help us - we need to see your work!
For instance, do you know the Law of sines and/or the Law of Cosines?

 

[soroban]

Hello, galfridus!

Where did this problem come from?
Some of the given information is wrong;
. . others are inaccurate.

Given all three side lengths and angles of a triangle, and two points,
find the position of the third point.

Points

. . \(\displaystyle \begin{array}{ccc}A &=& (0,0) \\ B&=& (5,15) \\ C &=& (?,?) \end{array}\)


Sides (length) . These are mislabeled.

. . \(\displaystyle \begin{array}{cccccc}AB\:\color{red}{(c)} &=& \sqrt{250} &\approx& 15.81 \\ AC\: (b) &=& 5 \\ BC\;\color{red}{(a)} &=& 14 \end{array}\)


Angles:

. . \(\displaystyle \begin{array}{ccc}A &=& 102^o \\ B &=& 18^o \\ C &=& 60^o \end{array}\) . These are only approximations.

You want a point \(\displaystyle C(x,y)\) so that:

. . \(\displaystyle \begin{array}{cccccccc}AC = 5: & (x-0)^2 + (y-0)^2 &=& 5^2 \\ BC = 14: & (x-5)^2 + (y-15)^2 &=& 14^2 \end{array}\)


Solve the system of equations.