$\mathrm{AB}$ 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 $\angle \mathrm{APB}$ ?

24°

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 \({24}^\circ\).