In sin(A - B) = $$\frac{1}{2}$$ and cos(A + B) =$$\frac{1}{2}$$ where A > B > 0 and A + B is an acute angle, then the value of B is?

Solution (By Examveda Team)

$$\eqalign{ & {\text{sin}}\left( {{\text{A}} - {\text{B}}} \right) = \frac{1}{2}{\text{ }}\left( {{\text{A}} - {\text{B}} = {\text{3}}{0^ \circ }} \right) \cr & {\text{cos}}\left( {{\text{A}} + {\text{B}}} \right) = \frac{1}{2}{\text{ }}\left( {{\text{A}} + {\text{B}} = {{60}^ \circ }} \right) \cr & {\text{Adding}}\,{\text{both}}\,{\text{side}} \cr & \Rightarrow \left( {{\text{A}} - {\text{B}}} \right) + \left( {{\text{A + B}}} \right) = {30^ \circ } + {60^ \circ } \cr & \Rightarrow 2{\text{A}} = {90^ \circ } \cr & \Rightarrow {\text{A}} = {45^ \circ } \cr & \because {\text{A}} - {\text{B}} = {30^ \circ } \cr & {\text{B}} = {\text{A}} - {30^ \circ } \cr & \Rightarrow {45^ \circ } - {30^ \circ } \cr & \Rightarrow {15^ \circ } \cr & \Leftrightarrow \frac{{15 \times \pi }}{{180}} = \frac{\pi }{{12}}({\text{radian}}) \cr} $$