# How do you find the exact value of the third side given triangle DEF, d=sqrt3, e=5, and mangleF=pi/6?

Dec 13, 2017

$\sqrt{13} \cong 3.61$

#### Explanation:

Since we don't know that the triangle is a right triangle, we're required to use the Cosine Law. I will also presume that $\angle F$ is opposite the unknown side.

The Law of Cosines states that:
${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \left(C\right)$, where $\angle C$ is the angle opposite side $c$.

In our case, it becomes (I'll call the unknown side $x$):
${x}^{2} = {5}^{2} + {\left(\sqrt{3}\right)}^{2} - 10 \sqrt{3} \cos \left(\frac{\pi}{6}\right)$

${x}^{2} = 25 + 3 - 10 \sqrt{3} \cos \left(\frac{\pi}{6}\right)$

${x}^{2} = 28 - 10 \sqrt{3} \cos \left(\frac{\pi}{6}\right)$

Next we take the square root on both sides:
$x = \sqrt{28 - 10 \sqrt{3} \cos \left(\frac{\pi}{6}\right)}$

$\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$

thereforex=sqrt(28-10*3/2

$x = \sqrt{28 - \frac{30}{2}} = \sqrt{28 - 15}$

$x = \sqrt{13}$