$\mathrm{AB}$ is a chord of a circle in minor segment with center $\mathrm{O} . \mathrm{C}$ is a point on the minor are of the circle between the points $\mathrm{A}$ and $\mathrm{B}$. The tangents to the circle at $\mathrm{A}$ and $\mathrm{B}$ meet at the point $\mathrm{P}$. If $\angle \mathrm{ACB}=102^{\circ}$, then what is the measure of $\angle \mathrm{APB}$ ?
Answer & explanation
Correct answer: option 4
Point D is taken on the major arc of the circle. Then , A & D, B&D, C & B, A & C are joined.
\(\angle\)ACB = \({102}^\circ\)
Now,
\(\angle\)ADB = (180 - 102) = \({78}^\circ\)
Now,
\(\angle\)AOB = (78 x 2) = \({156}^\circ\)
Since OA = OB, \(\Delta \)AOB is an isosceles triangle and \(\angle\)OBA = \(\angle\)OAB.
Now,
\(\angle\)AOB + \(\angle\)OBA + \(\angle\)OAB = \({180}^\circ\)
= 2 x \(\angle\)OBA = \({(180\; -\;156 )}^\circ\)
= \(\angle\)OBA = \({12}^\circ\)
Now,
\(\angle\)OBA = \({12}^\circ\)
\(\angle\)OAB = \({12}^\circ\)
Now,
\(\angle\)OAP = \(\angle\)OAB + \(\angle\)BAP
= \({90}^\circ\) = \({12}^\circ\) + \(\angle\)BAP
= \(\angle\)BAP = \({78}^\circ\)
Now,
\(\angle\)OBP = \(\angle\)OBA + \(\angle\)ABP
= \({90}^\circ\) = \({12}^\circ\) + \(\angle\)ABP
= \(\angle\)ABP = \({78}^\circ\)
Now,
\(\angle\)APB + \(\angle\)ABP + \(\angle\)BAP = \({180}^\circ\)
= \(\angle\)APB + \({78}^\circ\) + \({78}^\circ\) = \({180}^\circ\)
= \(\angle\)APB = \({180}^\circ\) - \({(2 \;×\; 78)}^\circ\)
= \(\angle\)APB = \({24}^\circ\)
Therefore, \(\angle\)APB is \({24}^\circ\).