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Question

# If $\mathrm{sin}\left(A-B\right)=\frac{1}{2}\mathrm{and}\mathrm{cos}\left(A+B\right)=\frac{1}{2},0°<\left(A+B\right)\le 90°\mathrm{and}A>B$ then values of A and B are (a) (60°, 30°) (b) (60°, 15°) (c) (45°, 15°) (d) (60°, 25°)

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Solution

## As we know that, $\mathrm{sin}30°=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\phantom{\rule{0ex}{0ex}}\mathrm{if}\mathrm{sin}\left(A-B\right)=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒A-B=30°...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{and}\mathrm{cos}60°=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\phantom{\rule{0ex}{0ex}}\mathrm{if}\mathrm{cos}\left(A+B\right)=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒A+B=60°...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Solving}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}A=45°\mathrm{and}B=15°\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{optiom}\mathrm{is}\left(\mathrm{c}\right).$

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