# If sides A and B of a triangle have lengths of 8 and 5 respectively, and the angle between them is (pi)/3, then what is the area of the triangle?

$A r e a = 10 \sqrt{3}$ square units

#### Explanation:

Use the formula $A r e a = \frac{1}{2} \cdot a b \sin C$ OR
$A r e a = \frac{1}{2} \cdot b c \sin A$ OR
$A r e a = \frac{1}{2} \cdot a c \sin B$
In this case side $a = 8$, side $b = 5$, angle $C = \frac{\pi}{3}$

$A r e a = \frac{1}{2} a b \sin C$
$A r e a = \frac{1}{2} \cdot 8 \cdot 5 \cdot \sin \left(\frac{\pi}{3}\right)$

$A r e a = \frac{1}{2} \cdot 8 \cdot 5 \cdot \frac{\sqrt{3}}{2}$

$A r e a = 10 \sqrt{3}$ square units

Jan 22, 2016

$10 \sqrt{3}$

#### Explanation:

In a triangle with 2 known sides , say a , and b , and the angle between them is $\theta$

then area (A) = $\frac{1}{2} a b \sin \theta$

In this question a = 8 , b = 5 and $\theta = \frac{\pi}{3}$

$\Rightarrow A = \frac{1}{2} \times 8 \times 5 \times \sin \left(\frac{\pi}{3}\right) = 20 \times \frac{\sqrt{3}}{2} = 10 \sqrt{3}$