Question

In $\Delta ABC$ if $\sin A\sin B=\frac{ab}{c^2},$ then $\angle C$ is

• A. ${90}^0$
• B. ${75}^0$
• C. ${60}^0$
• D. ${45}^0$

Answer 1

The correct option is A ${90}^0$

Given: $\sin A\sin B=\frac{ab}{c^2}\cdots \left(i\right)$
$\Rightarrow c^2=\frac{a}{\sin A}\cdot \frac{b}{\sin B}$
We know that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$
Putting in equation $\left(i\right)$
Thus $c^2=\frac{c^2}{{\sin }^{2}C}$
$\Rightarrow {\sin }^{2}C=1$
$\Rightarrow \angle C={90}^0$