waecmaths question:
In the diagram $NQ\parallel TS$ ,$\angle RTS={{50}^{\circ }}$ and $\angle PRT={{100}^{\circ }}$ Find the value of $\angle NPR$
Option A:
110o
Option B:
130o
Option C:
140o
Option D:
150o
waecmaths solution:
$\begin{align} & \angle QRP=180{}^\circ -\angle PRT\{Sum\text{ }of\text{ }angles\text{ }on\text{ }a\text{ }straight\text{ }line\} \\ & \angle QRP=180{}^\circ -100{}^\circ =80{}^\circ \\ & \angle PQT=\angle OTS=50{}^\circ \text{ }\{Alternate\text{ }angles\} \\ & \angle NPR=\angle PQT+\angle QRP\text{ }\!\!\{\!\!\text{ Sum of the two opp}\text{. interior angles in }\vartriangle =\text{opp}\text{. exterior angle }\!\!\}\!\!\text{ } \\ & \angle NPR=80{}^\circ +50{}^\circ =130{}^\circ \\\end{align}$
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