# Given the area and angles of an isosceles triangle, find the length of the sides.

An isosceles triangle has an area of $27 cm^2$, and the angle between the two equal sides is $\frac{5\pi}{6}$. What is the length of the two equal sides? (Round your answer to one decimal place.)

I don't even know where to start finding the answer for this. That's what I'm looking for.

• Area is given by $1/2r^2sin(5π/6)$ which is 27. Thus here two equal sides is r. Hence you get it. – Archis Welankar Nov 6 '15 at 16:56

Expounding on the above answer: You can use this formula for the area of a triangle for two sides and the opposite angle of opposite side, $$\text{Area}=\frac{1}{2}\text{ab}\sin \text{C}$$
Where $a$ and $b$ would be the lengths of the equal sides and $\text{C}$ is the angle between $a$ and $b$. But since $a$ and $b$ are equal in length let $a=c$ and $b=c$ so your formula becomes..$$\text{Area}=\frac{1}{2}\text{c}^2\sin \text{C}$$ Substitute the values you have and solve.
Letting $a$ be the length of the two equal sides, you'll have \begin{align}27=(1/2)a^2\sin(5\pi/6)&\iff a^2=\frac{27}{(1/2)\sin(5\pi/6)}\\&\iff a^2=\frac{27}{(1/2)\cdot (1/2)}\\&\iff a^2=108.\end{align} Since $a\gt0$, you have $$a=\sqrt{108}=\sqrt{6^2\cdot 3}=6\sqrt 3.$$