Question

With usual notation, if in a triangle $ABC$, $\cos A\cos B+\sin A\sin B\sin C=1$, then $a:b:c=x:x:y$. The value of $2\sqrt{2}y$ is

Answer 1

$\cos A\cos B+\sin A\sin B\sin C=1$
$\Rightarrow \sin C=\frac{1-\cos A\cos B}{\sin A\sin B}$
$\Rightarrow \frac{1-\cos A\cos B}{\sin A\sin B}\le 1$
$\Rightarrow 1-\cos A\cos B\le \sin A\sin B$
$\Rightarrow 1\le \cos (A-B)$
$\cos \theta$ cannot be greater than $1$
So, $\cos (A-B)=1$
$\Rightarrow A=B$

So, $\sin C=\frac{1-{\cos }^{2}A}{{\sin }^{2}A}$
$\Rightarrow \sin C=1\Rightarrow C=\pi /2$

Now, $A+B+C=\pi$
$\Rightarrow A=B=\frac{\pi }{4}$
$\sin A:\sin B:\sin C=\frac{1}{\sqrt{2}}:\frac{1}{\sqrt{2}}:1$
$\Rightarrow a:b:c=1:1:\sqrt{2}$

Hence, the value of $2\sqrt{2}y=2\sqrt{2}\times \sqrt{2}=4$