Question

In a $△ABC$, if $\sin A\sin B=\frac{ab}{c^2}$, then the triangle is

• A. Equilateral
• B. Isosceles
• C. Right-angled
• D. Obtuse-angle

Answer 1

The correct option is C

Right-angled

Determine the type of triangle.

We have, $\sin A\sin B=\frac{ab}{c^2}$

$$\begin{gathered} \Rightarrow c^2=\frac{ab}{\sin A\sin B} \\ \Rightarrow c^2=\left(\frac{a}{\sin A}\right)\left(\frac{b}{\sin B}\right)...\left(1\right) \end{gathered}$$

From sine rule we know that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$, therefore substitute $\frac{c}{\sin C}$ in equation $\left(1\right)$

$$\begin{gathered} \Rightarrow c^2=\frac{c^2}{{\sin }^{2}C} \\ \Rightarrow {\sin }^{2}C=1 \\ \Rightarrow \sin C=\sin 90° \\ \Rightarrow C=90° \end{gathered}$$

Thus, the triangle is right-angled triangle.

Hence, (C) option is correct.