Question

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

Answer 1

We know that ABCD is a cyclic quadrilateral.
$$\begin{gathered} \mathrm{Hence},\angle B+\angle D=180° \\ \Rightarrow \angle B=180°-\angle D \\ \Rightarrow \angle B=180°-130° \\ \Rightarrow \angle B=50° \end{gathered}$$

$$\begin{gathered} \mathrm{In}∆ABC, \\ \angle ABC+\angle ACB+\angle BAC=180°(\text{A}\mathrm{ngle}-\mathrm{sum}\mathrm{property}) \\ \Rightarrow \angle BAC=180°-\angle ABC-\angle ACB \\ \Rightarrow \angle BAC=180°-50°-90°(\angle ACB=90°,\text{as}\mathrm{angle}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{semi}-\mathrm{circle}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{angle}) \\ \Rightarrow \angle BAC=40° \end{gathered}$$