# How do you find the exact value of the third side given triangle RST, RS=9, ST=9sqrt3, mangleS=(5pi)/6?

Jan 15, 2017

$= 9 \sqrt{7}$

#### Explanation:

You would apply the cosine theorem to calculate the third side:

$\overline{R T} = \sqrt{{\overline{R S}}^{2} + {\overline{S T}}^{2} - 2 \cdot \overline{R S} \cdot \overline{S T} \cdot \cos \hat{S}}$

$= \sqrt{{9}^{2} + {\left(9 \sqrt{3}\right)}^{2} - 2 \cdot 9 \cdot 9 \sqrt{3} \cdot \cos \left(\frac{5}{6} \pi\right)}$

=sqrt(81+243-cancel162^81sqrt(3)*(-sqrt(3)/cancel2)

$= \sqrt{324 + 243} = \sqrt{567} = 9 \sqrt{7}$