solve the triangle/with given parts

**colie2020** · April 19th 2010, 12:32 PM

alpha=42.5degree
beta=35.2degree
a=18.0

**skeeter** · April 19th 2010, 02:09 PM

Originally Posted by colie2020

alpha=42.5degree
beta=35.2degree
a=18.0

$\alpha + \beta + \gamma = 180^\circ$

find $\gamma$ , then use the sine law.

**camdenreslink** · April 19th 2010, 02:13 PM

Originally Posted by colie2020

alpha=42.5degree
beta=35.2degree
a=18.0

I'm assuming side A is directly opposite angle $\alpha$ , there is a side B opposite angle $\beta$ , and there is a hypotenuse opposite angle $\gamma$ .

If so...

The sum of the interior angles in a triangle equals 180, so:
$ \alpha + \beta + \gamma = 180 $

$ 42.5 + 35.2 + \gamma = 180 $

$ \gamma = 102.3 $

The Law of Sines:

$ \frac{18}{\sin (42.5)} = \frac{B}{\sin (35.2)} $

$ B = \frac{18\sin (35.2)}{\sin (42.5)} $

$ \frac{18}{\sin (42.5)} = \frac{C}{\sin (102.3)} $

$ C = \frac{18\sin (102.3)}{\sin (42.5)} $

Check out the wikipedia entry on the law of sines:
Law of sines - Wikipedia, the free encyclopedia

What this problem is asking is known as triangulation:
Triangulation - Wikipedia, the free encyclopedia

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