# The question is below ?

## If in a $\Delta A B C , \cos A \cos B + \sin A \sin B \sin C = 1$ then show that $a : b : c = 1 : 1 : \sqrt{2}$

Jun 5, 2018

Given
$\cos A \cos B + \sin A \sin B \sin C = 1$
$\implies \cos A \cos B + \sin A \sin B - \sin A \sin B + \sin A \sin B \sin C = 1$

$\implies \cos \left(A - B\right) - \sin A \sin B \left(1 - \sin C\right) = 1$

$\implies 1 - \cos \left(A - B\right) + \sin A \sin B \left(1 - \sin C\right) = 0$
$\implies 2 {\sin}^{2} \left(\frac{A - B}{2}\right) + \sin A \sin B \left(1 - \sin C\right) = 0$

Now in above relation the first term being squared quantity will be positive.In the second term A,B and C all are less than
${180}^{\circ}$ but greater than zero.
So sinA ,sinB and sinC all are positive and less than 1.So the 2nd term as a whole is positive.
But RHS=0.
It is only possible iff each term becomes zero.

When $2 {\sin}^{2} \left(\frac{A - B}{2}\right) = 0$
then$A = B$

and when 2nd term=0 then
$\sin A \sin B \left(1 - \sin C\right) = 0$

0< A and B <180
$\implies \sin A \ne 0 \mathmr{and} \sin B \ne 0$

So $1 - \sin C = 0 \implies C = \frac{\pi}{2}$

So in triangle ABC
$A = B \mathmr{and} C = \frac{\pi}{2} \to \text{the triangle is right angled and isosceles}$

Side $a = b \mathmr{and} \angle C = {90}^{\circ}$

So$c = \sqrt{{a}^{2} + {b}^{2}} = \sqrt{{a}^{2} + {a}^{2}} = \sqrt{2} a$

Hence $a : b : c = a : 2 a : \sqrt{2} a = 1 : 1 : \sqrt{2}$