Question

Sin(A+B)=12/13. SinB=5/13. Then sinA

Answer 1

Since sinB is given,  take out cosB.
cosB = 12/13
Now,
sin(A+B)=12/13
Þ sinA cosB +sinB cosA = 12/13
Þ sinA × 12/13 + 5/13 ×cosA = 12/13
Þ 1/13 (12 sinA + 5 cosA) = 12/13
1/13 cancels giving
Þ  12 sinA +5cosA = 12
Þ  5 cosA = 12 -12 sinA
Now write cos and sin in the form of b/h and p/h
Þ  5 b/h = 12 - 12 p/h
Þ  5 b/h = 12(1 - p/h)
Þ  5 b/h = 12 (h-p)/h
Þ  5b =  12 (h - p)
Þ  5√(h2 -p2) = 12 (h - p)
 Squaring 
Þ  25 (h2 - p2) = 144 (h-p)(h-p)
Þ  25 (h+p)(h - p) = 144 (h-p)(h-p)
Þ 25 (h+p) = 144(h-p)
Þ 25h + 25p = 144h -144p
Þ 25p +144p = 144h -25h
Þ 169p = 119h
Þ  p/h = 119/169
sinA = p/h = 119/169

There might be some easier ways to find this. But, I don't know that. I just tried to figure the answer myself.
Hope it helps!

Answer 2

I hope this helps you.

:)