Question

In a triangle $ABC,\cos A\cos B+\sin A\sin B\sin C=1,$ then $a:b:c$ is:

• A. $1:1:\sqrt{2}$
• B. $1:\sqrt{2}:1$
• C. $\sqrt{2}:1:1$
• D. $\sqrt{2}:\sqrt{2}:1$

Answer 1

The correct option is A $1:1:\sqrt{2}$

Given: $\cos A\cos B+\sin A\sin B\sin C=1$
$\sin C=\frac{1-\cos A\cos B}{\sin A\sin B}\cdots \left(i\right)$
We know
$\Rightarrow \frac{1-\cos A\cos B}{\sin A\sin B}\le 1(\because \sin C\le 1)$
$\Rightarrow 1-\cos A\cos B\le \sin A\sin B$
$\Rightarrow 1\le \cos (A-B)$
Taking $\cos (A-B)=1\Rightarrow A-B=0$
Putting $A=B$ in equation $\left(i\right)$
$\sin C=\frac{1-{\cos }^{2}A}{{\sin }^{2}A}$
$\sin C=1\Rightarrow C=\pi /2$
$A+B+C=\pi$
$A+A+\frac{\pi }{2}=\pi$
$A=B=\frac{\pi }{4}$
$\therefore a:b:c=\sin A:\sin B:\sin C=\sin \frac{\pi }{4}:\sin \frac{\pi }{4}:\sin \frac{\pi }{2}$
$\therefore a:b:c=1:1:\sqrt{2}$