Question
If $\sin(\text{A}-\text{B})=\frac{1}{2}$ and $\cos(\text{A}+\text{B})=\frac{1}{2},0^\circ<\text{A}+\text{B}\geq90^\circ,\text{A}<\text{B}$ find A and B.
✓
Answer
$\sin(\text{A}-\text{B})=\sin30^\circ\ \cos(\text{A}+\text{B})=\cos60^\circ$$\text{A}-\text{B}=30^\circ\dots(\text{i})$
$\text{A}+\text{B}=60^\circ\dots(\text{ii})$
Add (i) & (ii) we get,
$2\text{A}=90^\circ,\text{A}=45^\circ$
$\text{A}-\text{B}=30^\circ$
$45-\text{B}=30^\circ\text{ B}=45-30^\circ$
$\text{B}=15^\circ$
$\text{A}+\text{B}=60^\circ\dots(\text{ii})$
Add (i) & (ii) we get,
$2\text{A}=90^\circ,\text{A}=45^\circ$
$\text{A}-\text{B}=30^\circ$
$45-\text{B}=30^\circ\text{ B}=45-30^\circ$
$\text{B}=15^\circ$
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