# solve the triangle/with given parts

• Apr 19th 2010, 12:32 PM
colie2020
solve the triangle/with given parts
alpha=42.5degree
beta=35.2degree
a=18.0
• Apr 19th 2010, 02:09 PM
skeeter
Quote:

Originally Posted by colie2020
alpha=42.5degree
beta=35.2degree
a=18.0

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

find $\displaystyle \gamma$ , then use the sine law.
• Apr 19th 2010, 02:13 PM
Quote:

Originally Posted by colie2020
alpha=42.5degree
beta=35.2degree
a=18.0

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

If so...

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

$\displaystyle 42.5 + 35.2 + \gamma = 180$

$\displaystyle \gamma = 102.3$

The Law of Sines:

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

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

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

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

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