Question

In a $△ABC,\cos B\cos C+\sin B\sin C{\sin }^{2}A=1$.Then the triangle is

• A. Right-angled isosceles
• B. Isosceles where equal angles are greater than $\frac{\pi }{4}$
• C. Equilateral
• D. None of these

Answer 1

The correct option is A Right-angled isosceles

$\Rightarrow$$A+B+C=\pi$

$\therefore$ $A=\pi -(B+C)$

Given, ${\sin }^{2}A=\frac{1-\cos B\cos C}{\sin B\sin C}\le 1$

$\Rightarrow \cos (B-C)\ge 1$

$\therefore \cos (B-C)=1=\cos 0\therefore B=C$

$\therefore {\sin }^{2}A=\frac{1-{\cos }^{2}B}{{\sin }^{2}B}=1\Rightarrow \angle A={90}^0$

$\therefore$ Triangle is right-angled isosceles