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Question

# In the given figure, O is the centre of a circle and ∠ADC = 130°. Find ∠BAC.

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Solution

## We know that ABCD is a cyclic quadrilateral. $\mathrm{Hence},\angle B+\angle D=180°\phantom{\rule{0ex}{0ex}}⇒\angle B=180°-\angle D\phantom{\rule{0ex}{0ex}}⇒\angle B=180°-130°\phantom{\rule{0ex}{0ex}}⇒\angle B=50°$ $\mathrm{In}∆ABC,\phantom{\rule{0ex}{0ex}}\angle ABC+\angle ACB+\angle BAC=180°\left(\text{A}\mathrm{ngle}-\mathrm{sum}\mathrm{property}\right)\phantom{\rule{0ex}{0ex}}⇒\angle BAC=180°-\angle ABC-\angle ACB\phantom{\rule{0ex}{0ex}}⇒\angle BAC=180°-50°-90°\left(\angle ACB=90°,\text{as}\mathrm{angle}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{semi}-\mathrm{circle}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{angle}\right)\phantom{\rule{0ex}{0ex}}⇒\angle BAC=40°\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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