Question

if angle A= angle B then in traingle ABC find value of-. sinA cosC + cosA sinC​

Answer 1

Step-by-step explanation:

In the triangle ABC . A+B+C=π . So, A=π−(B+C) . So, sinA=sin(π−(B+C))=sinπcos(B+C)−cosπsin(B+C)=sin(B+C)=sinBcosC+cosBsinC .

We are given that this is equal to 2sinBcosC .

Hence 2sinBcosC=sinBcosC+cosBsinC⟹sinBcosC=cosBsinC ,

Or sinBcosC−cosBsinC=sin(B−C)=0 .

As B, C are angles of a triangle, A,B,C>0 and A+B+C=π . So, the solution for sin(B−C)=0 cannot be of the form ±2nπ where n=1,2,… . B−C has to be 0. So, B=C which means ABC is an isosceles triangle.