Question

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

Answer 1

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

Explanation:

As `sinAsinB+cosAcosBsinC=1`

`2sinAsinB+2cosAcosBsinC=2=sin^2A+cos^2A+sin^2B+cos^2B`

i.e. `sin^2A+sin^2B-2sinAsinB+cos^2A+cos^2B-2cosAcosB+2cosAcosB-2cosAcosBsinC`

i.e. `(sinA-sinB)^2+(cosA-cosB)^2+2cosAcosB(1-sinC)=0`

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

`sinA-sinB=0`, `cosA-cosB=0` and `1-sinC=0`

• (the latter is true as `cosA` and `cosB` cannot be `0`)

Hence `A=B` and `sinC=1` i.e. `C=90^@`

Hence the triangle is an isosceles and right angled triangle.