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Question

# (i) If cos(40$°$ + x) = sin 30$°$ find the value of x. (ii) If tan y = sin 45$°$ + cos 45$°$ then find the value of y.

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Solution

## (i) $\mathrm{Given}:\mathrm{cos}\left(40°+x\right)=\mathrm{sin}30°\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know}\mathrm{that}\mathrm{cos}\mathit{}x°=\mathrm{sin}\left(90°-x\right)\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{cos}\left(40°+x\right)=\mathrm{sin}\left(90°-40°-x\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\left(40°+x\right)=\mathrm{sin}\left(50°-x\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}30°=\mathrm{sin}\left(50°-x\right)\left(\mathrm{Given}\right)\phantom{\rule{0ex}{0ex}}⇒30°=\left(50°-x\right)\phantom{\rule{0ex}{0ex}}⇒x=20°$ (ii) $\mathrm{Given}:\mathrm{tan}y=\mathrm{sin}45°\mathrm{cos}45°+\mathrm{sin}30°\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}y=\frac{1}{\sqrt{2}}×\frac{1}{\sqrt{2}}+\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}y=\frac{1}{2}+\frac{1}{2}=1\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}y=\mathrm{tan}45°\phantom{\rule{0ex}{0ex}}⇒y=45°\phantom{\rule{0ex}{0ex}}$

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