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:
Answer & explanation
Correct answer: option 2
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 \({16}^\circ\).