In a triangle A B C , if sin A sin B=(a b)/(c^2) , then the triangle is

Correct answer: (C) right angled. Given, sin A sin B=(a b)/(c^2) ⇒ c^2=(a b)/(sin A sin B)= (a/(sin A) ) (b/(sin B) ) ⇒ c^2= (c/(sin C) )^2 [∵ a/(sin A)=b/(sin B)=c/(sin C) ] ⇒ sin^2 C =1 ⇒ C =90^° Hence, triangle A B C is a right angled triangle.

Q98 / 170 · Manipal (MET) 2016 · Mathematics

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

A
equilateral
B
isosceles
Correct answer
C
right angled
D
obtuse angled

Solution

Given, $\sin A \sin B=\frac{a b}{c^2}$ $\begin{aligned} & \Rightarrow c^2=\frac{a b}{\sin A \sin B}=\left(\frac{a}{\sin A}\right)\left(\frac{b}{\sin B}\right) \\ & \Rightarrow c^2=\left(\frac{c}{\sin C}\right)^2 \quad\left[\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\right]\end{aligned}$ $\begin{aligned} \Rightarrow & & \sin ^2 C & =1 \\ \Rightarrow & & C & =90^{\circ}\end{aligned}$ Hence, $\triangle A B C$ is a right angled triangle.

Hence, the correct answer is option: 'C'.

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

Solution

More Properties of Triangles Questions