Question

If sin (A + B) = 1 and cos (A − B) = $\frac{\sqrt{3}}{2}$, where A > B, then B = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer 1

(a) 30^o

$$\begin{gathered} \sin (A+B)=1=\sin {90}^{\mathrm{o}}[\because \sin {90}^{\mathrm{o}}=1] \\ \Rightarrow A+B={90}^{\mathrm{o}}...\left(\mathrm{i}\right) \\ \cos (A-B)=\frac{\sqrt{3}}{2}=\cos {30}^{\mathrm{o}}\left[\because \cos {30}^{\mathrm{o}}=\frac{\sqrt{3}}{2}\right] \\ \Rightarrow A-B={30}^{\mathrm{o}}...\left(\mathrm{ii}\right) \\ \mathrm{Solving}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\mathrm{we}\mathrm{get}: \\ \mathrm{A}={60}^{\mathrm{o}}\mathrm{and}\mathrm{B}={30}^{\mathrm{o}} \\ \therefore \mathrm{B}={30}^{\mathrm{o}} \end{gathered}$$