# Thread: Request for help with 2 triangles problem

1. ## Request for help with 2 triangles problem

I have been trying to solve a problem; originally I was thinking of it as 2 circles, but it is basically a problem of triangles, please see the attached file:I know the distance between A and B, the length of r and the angle a. When I draw it out there can only be 1 answer per set of variables, but I can't get a handle on how to calculate any of the other lengths of the triangles' sides.

If anyone can point me in the right direction I would be greatful.

2. Originally Posted by EckCop
I know the distance between A and B, the length of r and the angle a
can you post what these values are

but so far looks you use the Law of Sines

3. ## values

Hi bigwave,

The values are a=5degrees, AB=110, r=66.

Hope this helps.

4. Hello, EckCop!

Solve the triangle.

Code:
                        C
*
*      *
*             *   66
*                    *
*                           *
* 5°                               *
A *  *  *  *  *  *  *  *  *  *  *  *  *  *  * B
110

We will find angle $C$ . . . Note: $\angle C$ is obtuse.

Law of Sines: . $\frac{\sin C}{110} \:=\:\frac{\sin 5^o}{66} \quad\Rightarrow\quad \sin C \:=\:0.145259571$

. . $\angle C \;=\;8.35231067^o\,\text{ or }\,171.6476893^o$

. . $\boxed{\angle C \;\approx\;171.65^o}$

Hence: . $\angle B \:=\:180^o - 5^o - 171.65^o \quad\Rightarrow\quad \boxed{\angle B \;\approx\;3.35^o}$

Law of Sines: . $\frac{AC}{\sin3.35^o} \:=\:\frac{66}{\sin5^o} \quad\Rightarrow\quad AC \:=\:44.25095312$

. . $\boxed{AC \;\approx\;44.25}$

5. Thanks Soroban,

That is just what I needed.