Question

In triangle ABC, $\sin A\sin B+\sin B\sin C+\sin C\sin A=9/4$ and $a=2$, then the value of $\sqrt{3}\Delta ,$ where $\Delta ,$ is the area of triangle, is___________ .

• A. $3$
• B. $4$
• C. $9$
• D. None of these

Answer 1

The correct option is A $3$

We know in $\Delta ABC$, If $\sin A\sin B+\sin B\sin C+\sin C\sin A=9/4$
Then triangle is equilateral. Hence it's area $\Delta =\frac{\sqrt{3}}{4}{a}^2=\sqrt{3}$

$\therefore \sqrt{3}\Delta =\sqrt{3}\times \sqrt{3}=3$