Question

Suppose you are given that tan(theta)=3. The hypotenuse=9, so find area of triangle. Please Help!

Answer 1

http://openstudy.com/study#/updates/508c005de4b0d596c4610b21

Answer 2

thx @hartnn

Answer 3

\[h=9\text{ Sin}[\text{ArcTan}[3]]\text{=}\frac{27}{\sqrt{10}}\]\[b\text{ = }9\text{ Cos}[\text{ArcTan}[3]]\text{=}\frac{9}{\sqrt{10}}\]\[A=\frac{1}{2}\frac{9}{\sqrt{10}} \frac{27}{\sqrt{10}}=\frac{243}{20}=12.15 \]