alpha=42.5degree

beta=35.2degree

a=18.0

Printable View

- Apr 19th 2010, 12:32 PMcolie2020solve the triangle/with given parts
alpha=42.5degree

beta=35.2degree

a=18.0 - Apr 19th 2010, 02:09 PMskeeter
- Apr 19th 2010, 02:13 PMcamdenreslink
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:__

http://upload.wikimedia.org/math/f/8...2d8d0e9f9d.png

$\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