Question

in a traingle ABC prove that a sin
(b-c) + b sin (c-a) + c sin (a-b) =0​

Answer 1

Answer:

proved

Step-by-step explanation:

If angles in a triangle be 60°, 60° 60° then

a sin ( b-c) + b sin(c-a) +c sin ( a-b)

a sin (60°-60°) +b sin (60°-60°) +c sin(60°-60°)

a sin0° +b sin0° +c sin0° =0+0+0=0