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 $\cos (B-C)=\cos B.\cos C+\sin B.\sin C$.

Hence for the purpose of simplification, we consider ${\sin }^{2}A=1$

Hence

$\cos \left(B\right).\cos \left(C\right)+\sin \left(B\right).\sin \left(C\right)=1$

$\cos (B-C)=\cos \left(0^0\right)$

$B=C$.

Since we had considered ${\sin }^{2}A=1$,

$A={90}^0$.

Hence, we got $A={90}^0$ and rest of the two angles as equal.

Thus the criteria for right angled isosceles triangle is fulfilled.