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can someone solve the triange im really stuck on it

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- Jul 25th 2007, 06:54 PMusm_67Trig Question
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can someone solve the triange im really stuck on it - Jul 25th 2007, 07:03 PMtukeywilliams
use law of cosines/ law of sines.

- Jul 25th 2007, 07:04 PMusm_67
well i know that since the chapter is based on it i just dont kno where to start

- Jul 25th 2007, 07:16 PMtukeywilliams
Is that $\displaystyle <AXB $?

- Jul 25th 2007, 07:17 PMusm_67
ye my bad its <axb

- Jul 25th 2007, 07:27 PMtukeywilliams
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 PMusm_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 PMtukeywilliams
The angles refer to the opposite sides. So $\displaystyle <XAB $ refers to side $\displaystyle BX $.