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

1 Answer
Feb 18, 2017

Answer:

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.