Question

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

• A. isosceles
• B. right angled
• C. equilateral
• D. right angle isosceles

Answer 1

Answer: (D)

right angle isosceles

If in a $\Delta ABC,\cos A.\cos B+\sin A.\sin B.\sin C=1$
$\Rightarrow \sin C=\frac{1-\cos A\cos B}{\sin A\sin B}\le 1$
$(\because max(\sin Q)=1,\forall Q\in R)$
$\Rightarrow \frac{1-\cos A\cos B}{\sin A\sin B}\le 1$
$\Rightarrow 1-\cos A\cos B-\sin A\sin B\ne 0$
$\Rightarrow 1-\cos A\cos B-\sin A\sin B\ne 0$
$\Rightarrow \cos (A-B)\ge 1$
Hence, only possible is $\cos (A-B)=1\Rightarrow A=B$
$\therefore$ The given relation reduce to $\sin C=\frac{1-{\cos }^{2}A}{{\sin }^{2}A}=1$
$\Rightarrow C={90}^0$.