# How do you find the area of triangle ABC given b=3, c=9, sinA=1/4?

##### 1 Answer
Dec 4, 2016

3.375 square units

#### Explanation:

$A r e a \left(\triangle A B C\right) = \frac{1}{2} b c \sin A$

= $\frac{1}{2} \cdot 3 \cdot 9 \cdot \frac{1}{4}$

= $\frac{27}{8}$

=$3 \frac{3}{8}$ or $3.375$

A (triangle ABC)= $3.375$ square units