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So I have triangle ABC where:

  1. $AC = BC$;
  2. $AB$ is known
  3. $\hat C$ (the angle $A\hat CB$) is known

I'm trying to find the altitude of said triangle.

MickG
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AlienNova
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  • The answer is $\frac{AB}{2\tan\frac{\angle C}{2}}$. Hint: the altitude from $C$ to $AB$ is also the angle bisector of $\angle C$ and the midpoint of $AB$. – user236182 Jan 05 '16 at 20:45

1 Answers1

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AB/2tan(∠C/2) should be the answer! The altitude is the angle bisector

Neil Philip
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