The vertices of a $\triangle A B C$ lie on a circle with centre $O$. $A O$ is produced to meet the circle at the point $P . D$ is a point on $\mathrm{BC}$ such that $\mathrm{AD} \perp \mathrm{BC}$. If $\angle \mathrm{B}=68^{\circ}$ and $\angle \mathrm{C}=52^{\circ}$, then the measure of $\angle \mathrm{DAP}$ is:

16°

Join O to B where point O is the circum center of the triangle.

= \(\angle\)BOA = 2\(\angle\)BCA

= 2 x 52 = 104

In triangle BOA

= OB = OA  [Radius of the circle]

= \(\angle\)OBA = \(\angle\)OAB

= \(\angle\)AOB + \(\angle\)OAB + \(\angle\)OBA = 180

= 2\(\angle\)OAB = (180 - 104)

= 2\(\angle\)OAB = 76

= \(\angle\)OAB = 38

In triangle

= \(\angle\)ABD + \(\angle\)BDA + \(\angle\)DAB = 180

= \(\angle\)DAB = 180 - 90 - 68

= \(\angle\)DAB = 22

Now,

\(\angle\)DAP = \(\angle\)BAO - \(\angle\)BAD

= \(\angle\)DAP = 38 - 22

= \(\angle\)DAP = \({16}^\circ\)

Therefore, \(\angle\)DAP is \({16}^\circ\).