# Request for help with 2 triangles problem

• May 28th 2010, 02:35 PM
EckCop
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:http://www.roughmustard.com/problem.bmpI 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.
• May 28th 2010, 03:33 PM
bigwave
Quote:

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
• May 29th 2010, 01:06 AM
EckCop
values
Hi bigwave,

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

Hope this helps.
• May 29th 2010, 05:49 AM
Soroban
Hello, EckCop!

Quote:

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

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

• May 29th 2010, 01:03 PM
EckCop
Thanks Soroban,

That is just what I needed.