MAT Exam 2022 Paper

Concept:Angle in a semicircle is $90^\circ$, angles on a straight line are supplementary, and vertically opposite angles are equal.Explanation:In the figure, $A$, $D$, $E$ lie on a straight line.Therefore, $\angle ADP + \angle PDE = 180^\circ$.Given $\angle ADP = 100^\circ$, we get $\angle PDE = 80^\circ$.Since $P$ lies on the circle and $DE$ is the diameter, $\angle DPE = 90^\circ$.In $\triangle PDE$,$\angle PED = 180^\circ - 90^\circ - 80^\circ = 10^\circ$.Also, $\angle BEQ = 100^\circ$, and $B,E,D$ are collinear.So, $\angle QED = 180^\circ - 100^\circ = 80^\circ$.Since $E,P,R$ are collinear and ray $EP$ lies between $EQ$ and $ED$,$\angle QER = \angle QED - \angle PED = 80^\circ - 10^\circ = 70^\circ$.Since $DE$ is a diameter and $Q$ lies on the circle, $\angle DQE = 90^\circ$.As $D,R,Q$ are collinear, $\angle RQE = 90^\circ$.In $\triangle QRE$,$\angle QRE = 180^\circ - 90^\circ - 70^\circ = 20^\circ$.The lines $PE$ and $QD$ intersect at $R$; hence, $\angle PRD$ and $\angle QRE$ are vertically opposite angles.Thus, $\angle PRD = \angle QRE = 20^\circ$.Answer:$\angle PRD = 20^\circ$, i.e., option A.