Question

If sin (A − B) = $\frac{1}{2}$ and cos (A + B) = $\frac{1}{2}$, 0° < (A + B) < 90° and A > B, then find A and B.

Answer 1

Here, sin (A − B) = $\frac{1}{2}$
⇒ sin (A − B) = sin 30^o [∵ sin 30^o = $\frac{1}{2}$]
⇒ A − B = 30^o​ ...(i)

Also, cos (A + B) = $\frac{1}{2}$
⇒​ cos (A + B) = cos 60^o [∵​ cos 60^o = $\frac{1}{2}$]
⇒ A + B = 60^o ...(ii)

Solving (i) and (ii), we get:
A = 45^o and B = 15^o