Question

If $\sin \left(A-B\right)=\frac{1}{2}\mathrm{and}\cos \left(A+B\right)=\frac{1}{2},0°<\left(A+B\right)\le 90°\mathrm{and}A>B$ then values of A and B are
(a) (60°, 30°)
(b) (60°, 15°)
(c) (45°, 15°)
(d) (60°, 25°)

Answer 1

As we know that,
$$\begin{gathered} \sin 30°=\frac{1}{2} \\ \mathrm{Thus}, \\ \mathrm{if}\sin \left(A-B\right)=\frac{1}{2} \\ \Rightarrow A-B=30°...\left(1\right) \\ \mathrm{and}\cos 60°=\frac{1}{2} \\ \mathrm{Thus}, \\ \mathrm{if}\cos \left(A+B\right)=\frac{1}{2} \\ \Rightarrow A+B=60°...\left(2\right) \\ \mathrm{Solving}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ A=45°\mathrm{and}B=15° \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{optiom}\mathrm{is}\left(\mathrm{c}\right). \end{gathered}$$