# Trig Question

• Jul 25th 2007, 06:54 PM
usm_67
Trig Question
http://img209.imageshack.us/my.php?image=new2zy6.jpg

can someone solve the triange im really stuck on it
• Jul 25th 2007, 07:03 PM
tukeywilliams
use law of cosines/ law of sines.
• Jul 25th 2007, 07:04 PM
usm_67
well i know that since the chapter is based on it i just dont kno where to start
• Jul 25th 2007, 07:16 PM
tukeywilliams
Is that $\displaystyle <AXB$?
• Jul 25th 2007, 07:17 PM
usm_67
• Jul 25th 2007, 07:27 PM
tukeywilliams
So $\displaystyle \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.

$\displaystyle \frac{50}{\sin 81.65} = \frac{b}{\sin 60}$. Solve for $\displaystyle b$. Then you can solve for the height from this information. So $\displaystyle \frac{h}{\sin 42.6} = \frac{b}{\sin 47.4}$. We already know $\displaystyle b$. So solve for $\displaystyle h$. So $\displaystyle h = \frac{b \sin 42.6}{\sin 47.4}$ or $\displaystyle h \approx 3.60b = 1748.661$.
• Jul 25th 2007, 07:35 PM
usm_67
i get everythin but how du u know what angles to use

like if its <XAB which angle is it referrin too

a rite
• Jul 25th 2007, 07:39 PM
tukeywilliams
The angles refer to the opposite sides. So $\displaystyle <XAB$ refers to side $\displaystyle BX$.