Question

Given that the sum of the angles A, B and C of a triangle is π radians, show that:
a) sin A = sin(B+C)
b) sin A + sin B + sin C = sin(A+B) + sin(B+C) + sin(A+C)

Please show all working clearly. Medal will be awarded.

Answer 1

\[A+B+C=\pi ,B+C=\pi-A\]
\[\sin \left( B+C \right)=\sin \left( \pi-A \right)=\sin A\]
b.
\[A+B+C=\pi,A+B=\pi-C,B+C=\pi-A,A+C=\pi-B\]
Solve as a.

Answer 2

I'm not quite understanding b

Answer 3

from a.you find sin(B+C),similarly find sin(C+A),And Sin (A+B) and then add three.

Answer 4

Can you tell me how you moved from sin(pi-a) to sina?

Answer 5

I am still not getting b :(

Answer 6

did you get the 1st part yet? a)

Answer 7

No I havent @jdoe0001

Answer 8

\[\sin \left( \pi-A \right)=\sin \pi \cos A-\cos \pi \sin A=0\times \cos A-\left( -1 \right)\sin A=\sin A\]

Answer 9

Oh, now I get that part

Answer 10

ahemmm ok

Answer 11

the idea being, the reference angle

Answer 12

Can you show it in steps ?

Answer 13

hmmm for ... part a) ?

Answer 14

For part b