BECE Mathematics 2014 Paper

Concept:Use the sum of angles in a triangle ($180^{\circ}$) and angles on a straight line ($180^{\circ}$) to find the unknown angle.Explanation:In $\triangle PQR$, the given angles are $50^{\circ}$ and $38^{\circ}$.Let $\angle PRQ = x$.Then $50^{\circ} + 38^{\circ} + x = 180^{\circ}$.So $x = 180^{\circ} - 88^{\circ} = 92^{\circ}$.This angle at $R$ is vertically opposite to the angle marked $e$, hence $e = 92^{\circ}$.On the straight line at $T$, we have $132^{\circ} + \angle RTS = 180^{\circ}$.Therefore $\angle RTS = 180^{\circ} - 132^{\circ} = 48^{\circ}$.In $\triangle RST$, the three angles are $e$, $d$, and $\angle RTS$.So $e + d + \angle RTS = 180^{\circ}$.Substitute $e = 92^{\circ}$ and $\angle RTS = 48^{\circ}$:$92^{\circ} + d + 48^{\circ} = 180^{\circ}$.$d + 140^{\circ} = 180^{\circ}$.$d = 180^{\circ} - 140^{\circ} = 40^{\circ}$.Answer:The angle marked $d$ is $40^{\circ}$, which is option B.