Question

# In the following figure, a circle with centre O is drawn and ∠BAC = 50°. Find x.

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Solution

## $\mathrm{As},\angle \mathrm{BOC}=2\angle \mathrm{BAC}\left(\mathrm{Angle}\mathrm{subtended}\mathrm{by}\mathrm{an}\mathrm{arc}\mathrm{at}\mathrm{the}\mathrm{centre}\mathrm{is}\mathrm{double}\mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{by}\mathrm{the}\mathrm{the}\mathrm{same}\mathrm{arc}\mathrm{at}\mathrm{any}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{circle}\right)\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{BOC}=2×50°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{BOC}=100°\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{in}∆\mathrm{BOC},\phantom{\rule{0ex}{0ex}}\therefore \mathrm{OB}=\mathrm{OC}\left(\mathrm{Radii}\right)\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{OBC}=\angle \mathrm{OCB}\left(\mathrm{Angles}\mathrm{opposite}\mathrm{to}\mathrm{equal}\mathrm{sides}\mathrm{are}\mathrm{equal}\right)\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{OBC}=\angle \mathrm{OCB}=x\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Also},\phantom{\rule{0ex}{0ex}}\angle \mathrm{BOC}+\angle \mathrm{OBC}+\angle \mathrm{OCB}=180°\left(\mathrm{Angle}\mathrm{sum}\mathrm{property}\mathrm{of}\mathrm{a}\mathrm{triangle}\right)\phantom{\rule{0ex}{0ex}}⇒100°+x+x=180°\phantom{\rule{0ex}{0ex}}⇒2x=180°-100°\phantom{\rule{0ex}{0ex}}⇒2x=80°\phantom{\rule{0ex}{0ex}}⇒x=\frac{80°}{2}\phantom{\rule{0ex}{0ex}}\therefore x=40°$

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