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

Correct answer: (C) right angled. Hints : 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 ⇒ sin^2 C=1 ⇒ sin C=1 ⇒ C=90^°

Q51 / 160 · WBJEE 2009 · Mathematics

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

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

Solution

$$ \begin{aligned} & \text { Hints }: \sin A \sin B=\frac{a b}{c^2} \\ & \Rightarrow c^2=\frac{a b}{\sin A \sin B}=\left(\frac{a}{\sin A}\right)\left(\frac{b}{\sin B}\right) \\ & \Rightarrow c^2=\left(\frac{c}{\sin C}\right)^2 \Rightarrow \sin ^2 C=1 \Rightarrow \sin C=1 \Rightarrow C=90^{\circ} \end{aligned} $$

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

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

Solution

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