Question

If in a triangle $ABC$ $\cos A\cos B+\sin A\sin B\sin C=1$
Show that $a:b:c=1:1:\sqrt{2}$.

Answer 1

From the given relation

$\sin C=\frac{1-\cos A\cos B}{\sin A\sin B}\le 1$

$\Rightarrow 1\le \cos A\cos B+\sin A\sin B$

or $1\le \cos (A-B)$ or $\cos (A-B)\ge 1$

$\Rightarrow \cos (A-B)=1$ $\Rightarrow A-B=0$

$\therefore A=B\because \cos (A-B)\ngtr 1$

Hence from $\left(1\right)$

$\sin C=\frac{1-{\cos }^{2}A}{{\sin }^{2}A}=\frac{{\sin }^{2}A}{{\sin }^{2}A}=1$

$\therefore C={90}^o$ $\therefore A+B={90}^o$

or $A=B={45}^o$ by $\left(2\right)$

$a:b:c=\sin A:\sin B:\sin C$

$=\frac{1}{\sqrt{2}}:\frac{1}{\sqrt{2}}:1=1:1:\sqrt{2}$