Question

In $△ABC$, if $\cos B\cos C+\sin B\sin C{\sin }^{2}A=1$, then

• A. $\angle A={45}^{\circ }$
• B. $\angle A={60}^{\circ }$
• C. $\angle B=\angle C$
• D. $\angle A={90}^{\circ }$

Answer 1

Answer: (C), (D)

The correct options are
C $\angle B=\angle C$
D $\angle A={90}^{\circ }$

$\cos B\cos C+\sin B\sin C{\sin }^{2}A=1...\left(1\right)$

$\sin B\sin C{\sin }^{2}A\le \sin B\sin C(\because {\sin }^{2}A\le 1)$
$\Rightarrow \cos B\cos C+\sin B\sin C{\sin }^{2}A\le \cos B\cos C+\sin B\sin C$
$\Rightarrow \cos B\cos C+\sin B\sin C{\sin }^{2}A\le \cos (B-C)$
$\Rightarrow 1\le \cos (B-C)$
$\Rightarrow \cos (B-C)=1[\because max(\cos x)=1]$
$\Rightarrow B=C$

From equation $\left(1\right)$,
${\cos }^{2}B+{\sin }^{2}B{\sin }^{2}A=1$
$\Rightarrow {\sin }^{2}B{\sin }^{2}A={\sin }^{2}B$
$\Rightarrow {\sin }^{2}A=1$
$\Rightarrow \angle A={90}^{\circ }$