Question

If $\sin (A-B)=\frac{1}{2},\cos (A+B)=\frac{1}{2},0^o<A+B\le {90}^o,A>B$, find $A$ and $B$.

Answer 1

We know that $sin30=\frac{1}{2}$ and $cos60=\frac{1}{2}$

$A-B=30,A+B=60$, Therefore, $2A=90,A={45}^o$,

$45+B=60$, $B={15}^o$