Question

If in a triangle $ABC,\cos A\cos B+\sin A\sin B\sin C=1,$ then the sides are proportional to

• A. $1:1:\sqrt{2}$
• B. $1:\sqrt{2}:1$
• C. $\sqrt{2}:1:1$
• D. none of these

Answer 1

Answer: (A)

$1:1:\sqrt{2}$

$\cos A\cos B+\sin A\sin B\sin C=1$
$\Rightarrow \sin A\sin B\sin C=1-\cos A\cos B$
$\Rightarrow \sin C=\frac{1-\cos A\cos B}{\sin A\sin B}\le 1$ .............$\left(1\right)$
$\Rightarrow 1\le \cos A\cos B+\sin A\sin B\sin C$
or $\cos (A-B)\ge 1$
$\therefore \cos (A-B)=1=\cos 0$
$\Rightarrow A-B=0$ or $A=B$
Eqn$\left(1\right)\Rightarrow \sin C=\frac{1-{\cos }^{2}C}{{\sin }^{2}A}=1$
$\therefore \angle C={90}^0$
$\therefore \angle A=\angle B={45}^0$
$a:b:c=\sin A:\sin B:\sin C$
$=\sin {45}^0:\sin {45}^0:\sin {90}^0$
$=\frac{1}{\sqrt{2}}:\frac{1}{\sqrt{2}}:1$
$=1:1:\sqrt{2}$