# If sides A and B of a triangle have lengths of 2 and 9 respectively, and the angle between them is (5pi)/6, then what is the area of the triangle?

Aug 27, 2016

4.5 square units.

#### Explanation:

Given a triangle, where 2 sides and the angle between them are known, as in this question. Then we can calculate the area (A) of the triangle using.

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{A = \frac{1}{2} \times a \times b \times \sin \left(\text{angle between them}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where a and b are the 2 known sides.

here a = 2 , b = 9 and angle between them $= \frac{5 \pi}{6}$

$\Rightarrow A = \frac{1}{2} \times 2 \times 9 \times \sin \left(\frac{5 \pi}{6}\right)$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\sin \left(\frac{5 \pi}{6}\right) = \sin \left(\pi - \frac{5 \pi}{6}\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\Rightarrow A = \frac{1}{2} \times 2 \times 9 \times \sin \left(\frac{\pi}{6}\right)$

$\Rightarrow A = \frac{1}{2} \times 2 \times 9 \times \frac{1}{2} = 4.5 \text{ square units}$