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The basic triangle looks something like this:

triangle 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.



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See this – Apurv Feb 27 '14 at 4:26
Thanks! I think that might have already solved the problem =) – evamvid Feb 27 '14 at 4:31
Nope =( In my problem, the numbers didn't work quite as nicely out (it's hard to explain why -- it had to do with which sides and angles were given) – evamvid Feb 27 '14 at 4:36
You wrote "oblique" triangle. The one in your question isn't oblique but acute. – DonAntonio Feb 27 '14 at 4:55
@evamvid, you can alternatively use the cosine rule. See my answer below – Apurv Feb 27 '14 at 5:02

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 $$h=a\sin38^\circ \implies h=41.265951, 2.3973563 $$

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When you say that $a=67.02701552,3.89395228$, what does the comma do? I saw this in an answer on a similar question a couple of days ago and I still don't understand. – evamvid Feb 27 '14 at 12:03
@evamvid, It means that two values for $a$, and consequently, two triangles exist. – Apurv Feb 28 '14 at 4:15

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