Question

If in a triangle $ABC$, $\cos A\cos B+\sin A\sin B\sin C=1$, then the triangle is

• A. Isosceles
• B. Right angled
• C. Equilateral
• D. none of these

Answer 1

Answer: (B)

Right angled

Given, $\cos A.\cos B+\sin A.\sin B.\sin C=1$

Multiply both sides by $2$, we get

$2\cos A\cos B+\sin A\sin B\sin C=({\cos }^{2}A+{\sin }^{2}A)+({\cos }^{2}B+{\sin }^{2}B)$

$\Rightarrow {\cos }^{2}A+{\cos }^{2}B-2\cos A\cos B+{\sin }^{2}A+{\sin }^{2}B-2\sin A\sin B+2\sin A\sin B-2\sin A\sin B\sin C=0$

$(\cos A-\cos B{)}^2+(\sin A-\sin B{)}^2+2\sin A\sin B(1-\sin C)=0$

$\cos A=\cos B,\sin A=\sin B⟹A=B$

and $(1-\sin C)=0⟹\sin C=1$

Therefore, $△ABC\text{is right angled isosceles triangle}$.