MAT Exam 2022 Paper

Concept:Use the angle in a semicircle, exterior angle property of a triangle, and angles in the same segment.Explanation:AD is the diameter of the circle.So, the angle subtended by AD at point B is:$$\angle \mathrm{ABD} = 90^\circ$$Since PBA is a straight line:$$\angle \mathrm{DBP} + \angle \mathrm{ABD} = 180^\circ$$$$\angle \mathrm{DBP} = 180^\circ - 90^\circ = 90^\circ$$In $\triangle \mathrm{BDP}$:$$\angle \mathrm{BDP} = 180^\circ - (\angle \mathrm{DBP} + \angle \mathrm{P})$$$$\angle \mathrm{BDP} = 180^\circ - (90^\circ + 20^\circ) = 70^\circ$$Since PDC is a straight line, $\angle \mathrm{BDP}$ is the exterior angle of $\triangle \mathrm{BDC}$:$$\angle \mathrm{BDP} = \angle \mathrm{DCB} + \angle \mathrm{DBC}$$Angles $\angle \mathrm{DCB}$ and $\angle \mathrm{DAB}$ are in the same segment of chord DB:$$\angle \mathrm{DCB} = \angle \mathrm{DAB} = 40^\circ$$Therefore:$$\angle \mathrm{DBC} = 70^\circ - 40^\circ = 30^\circ$$Answer:Option A: $30^\circ$