AB is a chord of a circle with centre O, while PAQ is the tangent at A. R is a point on the mirror arc AB. If ∠BAQ = 70^o, then find the measure of ∠ARB.

110^o

PAQ is tangent at point A i.e; \(\angle\)OAQ = \({90}^\circ\)

From this figure, \(\angle\)OAB = \(\angle\)OAQ - \(\angle\)BAQ

⇒ \(\angle\)OAB = \({90}^\circ\) - \({70}^\circ\) = \({20}^\circ\)

\(\angle\)OBA = \(\angle\)OAB = \({20}^\circ\) (since angle submitted by radius are equal)

In \(\Delta \)OBA, \(\angle\)BOA = \({180}^\circ\) - (\(\angle\)OBA + \(\angle\)OAB)

⇒ \(\angle\)BOA = \({180}^\circ\) - (\({20}^\circ\) - \({20}^\circ\)) = \({180}^\circ\) - \({40}^\circ\) = \({140}^\circ\)

\(\angle\)BOA (exterior angle) = \({360}^\circ\) - \({140}^\circ\) = \({220}^\circ\)

\(\angle\)ARB = \(\frac{1}{2}\)\(\angle\)BOA(exterior angle)

\(\angle\)ARB = \(\frac{1}{2}\) x \({220}^\circ\) = \({110}^\circ\)

Therefore, \(\angle\)ARB is \({110}^\circ\)