$\cos(a)=4/5$, $\sin(b)=12/13$, what is $\sin(c)$? with $a,b,c < \frac{\pi}{2}$.
Attempt :
Since $\sin^{2}(a) = 1 - \cos^{2}(a)$, I get $\sin(a) = \frac{3}{5}$. But how to get $\sin(c)$?
I have tried using $$ \frac{A}{\sin(a)} = \frac{B}{\sin(b)} = \frac{C}{\sin(c)}$$ and get $$ \frac{A}{B} = \frac{13}{20}= \frac{\sin(a)}{\sin(b)} $$ so $A=13, B = 20$ . How to find $\sin(c)$? Thanks.
Note that $$c=\pi -(a+b)$$ Therefore $$\sin c = \sin (a+b)=\sin a \cos b +\cos a \sin b=(3/5)(5/13)+(4/5)(12/13)=63/65 $$