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?
A. $$\frac{\pi }{6}$$
B. $$\frac{\pi }{{12}}$$
C. $$\frac{\pi }{4}$$
D. $$\frac{\pi }{2}$$
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & {\text{sin}}\left( {{\text{A}} - {\text{B}}} \right) = \frac{1}{2}{\text{ }}\left( {{\text{A}} - {\text{B}} = {\text{3}}{0^ \circ }} \right) \cr & {\text{cos}}\left( {{\text{A}} + {\text{B}}} \right) = \frac{1}{2}{\text{ }}\left( {{\text{A}} + {\text{B}} = {{60}^ \circ }} \right) \cr & {\text{Adding}}\,{\text{both}}\,{\text{side}} \cr & \Rightarrow \left( {{\text{A}} - {\text{B}}} \right) + \left( {{\text{A + B}}} \right) = {30^ \circ } + {60^ \circ } \cr & \Rightarrow 2{\text{A}} = {90^ \circ } \cr & \Rightarrow {\text{A}} = {45^ \circ } \cr & \because {\text{A}} - {\text{B}} = {30^ \circ } \cr & {\text{B}} = {\text{A}} - {30^ \circ } \cr & \Rightarrow {45^ \circ } - {30^ \circ } \cr & \Rightarrow {15^ \circ } \cr & \Leftrightarrow \frac{{15 \times \pi }}{{180}} = \frac{\pi }{{12}}({\text{radian}}) \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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