Question

If $\sin (A+B)=\frac{\sqrt{3}}{2}$ and $\sin (A-B)=\frac{1}{2}$ when 0 < A, B < ${90}^{\circ }$.
What is the value of A?

Answer 1

Given,
$\sin (A+B)=\frac{\sqrt{3}}{2}$
And,
$\sin (A-B)=\frac{1}{2}$

We know that,
$\sin {60}^{\circ }=\frac{\sqrt{3}}{2}$
So, $\sin {60}^{\circ }=\sin (A+B)$
On comparing, A + B = ${60}^{\circ }$ (i)

We also know that,
$\sin {30}^{\circ }=\frac{1}{2}$
So, $\sin {30}^{\circ }=\sin (A-B)$
On comparing, A - B = ${30}^{\circ }$ (ii)

Add equation (i) and equation (ii),
(A + B) + (A - B) = 60 + 30
A + B + A - B = 90
2A = 90
A = $\frac{{90}^{\circ }}{2}$
A = ${45}^{\circ }$

Thus, A = ${45}^{\circ }$