1. ## Trig Question

can someone solve the triange im really stuck on it

2. use law of cosines/ law of sines.

3. well i know that since the chapter is based on it i just dont kno where to start

4. Is that $\displaystyle <AXB$?

5. ye my bad its <axb

6. 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 $\displaystyle h \approx 3.60b = 1748.661$.

7. i get everythin but how du u know what angles to use

like if its <XAB which angle is it referrin too

a rite

8. The angles refer to the opposite sides. So $\displaystyle <XAB$ refers to side $\displaystyle BX$.