Question

If $\cos B\cos C+\sin B\sin C{\sin }^{2}A=1$, then triangle $ABC$ is

• A. isosceles and right angled
• B. equilateral
• C. isosceles whose equal angles are greater than $\pi /4$
• D. none

Answer 1

The correct option is A isosceles and right angled

We know that $\cos (B-C)=\cos B.\cos C+sinB.sinC$.
Hence for the purpose of simplification, we consider ${\sin }^{2}A=1$
Hence
$\cos \left(B\right).\cos \left(C\right)+\sin B.\sin C=1$
$cos(B-C)=cos\left(0^0\right)$
$B=C$.
Since we had considered ${\sin }^{2}A=1$,
$\therefore A={90}^0$.
Hence we got $A={90}^0$ and rest of the two angles as equal.
Thus the criteria for right angled isosceles triangle is fulfilled.
Hence answer is Option A.