# If sides A and B of a triangle have lengths of 1 and 2 respectively, and the angle between them is (2pi)/3, then what is the area of the triangle?

Jan 16, 2018

The area is $\frac{\sqrt{3}}{2}$; or, *0.866 *

#### Explanation:

You'll need the sine rule for area of the triangle.
The formula is:
Area of the $\triangle A B C = \frac{1}{2} \cdot a \cdot b \cdot \sin C$
This drawing(sorry if it's clumsy) will show what I mean :

Here, $\angle C = \frac{2 \pi}{3}$
$\sin C = \frac{\sqrt{3}}{2}$
Plugging in the values;
a=1;
b=2;
$\sin \left\{\frac{2 \pi}{3}\right\} = \frac{\sqrt{3}}{2}$
Area of triangle ABC = 1/2*1*2*sqrt3/2 =sqrt3/2
*Do not forget to give the units...