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

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Answer 1 · 2017-02-18T11:16:48

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

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 $1-sinC=0$

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

Hence the triangle is an isosceles and right angled triangle.

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