How do I solve for the height of a triangle?

Question

The basic triangle looks something like this:

How do I solve for $h$? As an example, in one problem I was given $b = 45, c = 42, \angle C = 38^\circ$

I understand how $h$ divides $\triangle ABC$ into two right triangles, and then you can find the upper angle (what used to be $\angle B$) of the right-hand triangle by using the the Triangle Angle Sum Theorem. But after that, I'm stuck. I'm pretty sure it has something to do with the Law of Sines, but I'm not sure what, exactly.

Thanks!

evamvid

Answer 1

Note that $$\dfrac {h}{a}=\sin 38^\circ$$ and using the cosine rule for triangle $ABC$, we have $$c=\sqrt {a^2+b^2-2ab\cos C}$$ or $$a^2-2ab\cos C+b^2-c^2=0$$ Plugging in values and solving for $a$, we get $$a=67.02701552,3.89395228$$ which gives $$h=a\sin38^\circ \implies h=41.265951, 2.3973563 $$