# If in a triangle ABC, sinAsinB+cosAcosBsinC=1 then how will you prove that the triangle is right angled and isosceles?

Feb 18, 2017

The triangle is an isosceles and right angled triangle. See proof below.

#### Explanation:

As $\sin A \sin B + \cos A \cos B \sin C = 1$

$2 \sin A \sin B + 2 \cos A \cos B \sin C = 2 = {\sin}^{2} A + {\cos}^{2} A + {\sin}^{2} B + {\cos}^{2} B$

i.e. ${\sin}^{2} A + {\sin}^{2} B - 2 \sin A \sin B + {\cos}^{2} A + {\cos}^{2} B - 2 \cos A \cos B + 2 \cos A \cos B - 2 \cos A \cos B \sin C$

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

Now the sum of the three terms on LHS will be $0$, only if

$\sin A - \sin B = 0$, $\cos A - \cos B = 0$ and $1 - \sin C = 0$

• (the latter is true as $\cos A$ and $\cos B$ cannot be $0$)

Hence $A = B$ and $\sin C = 1$ i.e. $C = {90}^{\circ}$

Hence the triangle is an isosceles and right angled triangle.