If $$\sin \alpha + \cos \beta = 2;$$   $$\left( {{0^ \circ } \leqslant \beta < \alpha \leqslant {{90}^ \circ }} \right){\text{,}}$$     then $${\text{sin}}\,{\left( {\frac{{2\alpha + \beta }}{3}} \right)^ \circ }$$   is?

Solution (By Examveda Team)

$$\eqalign{ & \sin {\text{ }}\alpha + \cos \beta = 2 \cr & {\bf{Shortcut\,\, method:}} \cr & {\text{Put, }}\alpha = {90^ \circ },\beta = {0^ \circ } \cr & \Leftrightarrow {\text{sin }}{90^ \circ } + {\text{cos }}{0^ \circ } = 2 \cr & \Leftrightarrow 1 + 1 = 2 \cr & \Leftrightarrow 2 = 2\left[ {{\text{Matched}}} \right] \cr & {\text{So, }}\alpha = {90^ \circ },\beta = {0^ \circ } \cr & \Rightarrow {\text{sin }}{\left( {\frac{{2\alpha + \beta }}{3}} \right)^ \circ }{\text{ }} \cr & \Rightarrow {\text{sin }}{\left( {\frac{{2 \times {{90}^ \circ } + {0^ \circ }}}{3}} \right)^ \circ } \cr & \Rightarrow {\text{sin }}{\left( {\frac{{{{180}^ \circ }}}{3}} \right)^ \circ } \cr & \Rightarrow {\text{sin }}{60^ \circ } = \cos {30^ \circ } = \frac{{\sqrt 3 }}{2} \cr & {\text{Take cos}}\frac{\alpha }{3} = \cos \frac{{{{90}^ \circ }}}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{cos 3}}{0^ \circ } \cr & {\text{So, this is answer }}{\text{.}} \cr} $$