solve this!!!!!!!?

Question

sin2a+sin2b+sin2c=-------------if a+b+c=180

Answer 1

answer = 4sinAsinBsinC

Answer 2

sin2a+sin2b+sin2c= {sin 2a + sin 2 b }+sin 2c
Using the formula sin C + sin D = 2 sin(C+D)/2 cos (C - D)/2
= {2 sin(2a+2b)/2 cos (2a - 2b)/2} + sin 2c
use the formula sin 2A= 2 sin A Cos A
= {2 sin (a + b) cos (a - b)} + 2sin c cos c
Since (a+ b + c) = 180
(a+ b) =180 - c
= {2 sin (180 - c) cos (a - b)} + 2sin c cos c
Since sin(180 - c) = sinc
={2 sin c cos (a - b)} + 2sin c cos c
=2sin c { cos (a - b) + cos c }
Sine( a+ b + c)= 180
Therefore c= 180 - (a + b )
= 2sin c { cos (a - b) + cos[180 -(a + b] }
Since cos[180 -(a + b] = - cos ( a + b )
= 2sin c { cos (a - b) - cos(a + b)}
Using the formula cos C - cos D = 2 sin ( C+ D)/2 sin (D - C)/2
= 2 sin c { 2 sin a sin b}
= 4 sin a sin b sin c
Hence
sin2a+sin2b+sin2c= 4 sin a sin b sin c

Answer 3

30

Answer 4

Hint: imagine a triangle inscribed in a unit circle. Use the inscribed-angle theorem.

Answer 5

uhhh, why don't you let a=b=c=60, then let a=b=90 and c=0 and see what you get.

Answer 6

what huh im confused

Answer 7

sin2a+sin2b+sin2c=2sin(a+b)cos(a-b)+2sinccosc
2sin[180-(c)]cos(a-b)+2sinccosc
2sinc[cos(a-b)+cosc]
2sinc*[cos(a-b)+cos(pi-(a+b)]
2sinc[cos(a-b)-cos(a+b)]
2sinc*2sinasinb
4sinasinbsinc