Question

Sollutions of triangle

Answer 1

I don't get how they get to this result

Answer 2

@sidsiddhartha @SithsAndGiggles

Answer 3

@sidsiddhartha

Answer 4

yeah and
A+B+C = pi

Answer 5

\[\huge A+B+C = \pi \]

Answer 6

Recall the double angle identity for sine. The numerator can be rewritten so that the denominator is canceled.

Answer 7

i think u gotta use some thing like\[sinA/2=\sqrt{\frac{ (s-b)(s-c) }{ bc }}\\sinB/2=\sqrt{\frac{ (s-c)(s-a) }{ ac }}\\sinC/2=\sqrt{\frac{ (s-a)(s-b) }{ ab} }\]

Answer 8

\[4\sin(A/2)\sin(B/2)\sin(C/2)=4.\frac{ (s-a)(s-b)(s-c) }{ abc }\]

Answer 9

genius ,

Answer 10

I was thinking more along the lines of
\[\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}\]

Answer 11

^Not to discredit sid's approach :)

Answer 12

thank you everyone =:)