Question
Evaluate the following:
If $\sin(\text{A}-\text{B})=\frac12$ and $\cos(\text{A}+\text{B})=\frac12,0^\circ<(\text{A}+\text{B})<90^\circ$ and $\text{A}>\text{B}$ then find A and B.

Answer

Here, $\sin(\text{A}-\text{B})=\frac12$
$\Rightarrow\sin(\text{A}-\text{B})=\sin30^\circ$ $\Big[\because\ \sin30^\circ=\frac12\Big]$
$\Rightarrow\text{A}-\text{B}=30^\circ\dots(\text{i})$
Also, $\cos(\text{A}+\text{B})=\frac12$
$\Rightarrow\cos(\text{A}+\text{B})=\cos60^\circ$ $\Big[\because\ \cos60^\circ=\frac12\Big]$
$\Rightarrow\text{A}+\text{B}=60^\circ\dots(\text{ii})$
Solving (i) and (ii), we get:
A = 45° and B = 15°.

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