Question

if sin A + sin B + sin C = 3,then find the value of sin(A + B + C) = ?​

Answer 1

Answer:

If sinA+sinB+sinC=3 , then cosA+cosB+cosC=?

sinA+sinB+sinB=1+1+1

or. (sinA-1)+(sinB-1)+(sinC-1)=0………..(1)

Comparing the both sides of eqn. (1)

(sinA-1)=0. => sinA= 1 or A=90°

(sinB-1)=0 => sinB=1 or B=90°

(sinC-1)=0. => sinC=1 or C=90°.

Now cosA+cosB+cosC=?

Putting. A=B=C=90°

=cos90°+cos90°+cos90°

=0+0+0

= 0. Answer

use these same steps to answer ur question

Answer 2

Answer:

We know that

sin 90 = 1

then let,

A,B,C = 90

sin A + sin B + sin C =3

then sin ( A+B+C) = sin ( 90 + 90 + 90 )

sin (270) = -1