Question

If $\sin (A+B)=1$ and $\cos (A-B)=\frac{\sqrt{3}}{2}$ Find A and B.

Answer 1

Given that.

$\sin (A+B)=1......\left(1\right)$

$\cos (A-B)=\frac{\sqrt{3}}{2}......\left(2\right)$

By equation (1) to,

$\sin (A+B)=1$

$\sin (A+B)=\sin {90}^0$

$A+B={90}^0......\left(3\right)$

By equation (2) to,

$\cos (A-B)=\frac{\sqrt{3}}{2}$

$\cos (A-B)=\cos {30}^0$

$A-B={30}^0....\left(4\right)$

From (1) and (2) to,

$A={60}^0$ and $B={30}^0$

Hence, this is the answer.