Question

If $\sin \left(A+B\right)=\frac{\sqrt{3}}{2}\mathrm{and}\cos \left(A-B\right)=\frac{\sqrt{3}}{2},0°<\left(A+B\right)\le 90°\mathrm{and}A>B$ then find the values of A and B.

Answer 1

As we know that,
$$\begin{gathered} \sin 60°=\frac{\sqrt{3}}{2} \\ \mathrm{Thus}, \\ \mathrm{if}\sin \left(A+B\right)=\frac{\sqrt{3}}{2} \\ \Rightarrow A+B=60°...\left(1\right) \\ \mathrm{and}\cos 30°=\frac{\sqrt{3}}{2} \\ \mathrm{Thus}, \\ \mathrm{if}\cos \left(A-B\right)=\frac{\sqrt{3}}{2} \\ \Rightarrow A-B=30°...\left(2\right) \\ \mathrm{Solving}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ A=45°\mathrm{and}B=15° \\ \mathrm{Hence},\mathrm{the}\mathrm{values}\mathrm{of}A\mathrm{and}B\mathrm{are}45°\mathrm{and}15°,\mathrm{respectively}. \end{gathered}$$