Question 23

loading...

Prev/Next links

waecmaths question: 

In the diagram $PR\parallel SV\parallel WY,\text{ }TX\parallel QY,\text{ }\angle TXW={{60}^{\circ }}$ find $\angle TQU$

 

Option A: 

120o

Option B: 

108o

Option C: 

72o

Option D: 

60o

waecmaths solution: 

$\begin{align}  & \angle QTU=\angle PQT={{48}^{\circ }}\text{    }\!\!\{\!\!\text{ alternate angles }\!\!\}\!\!\text{ } \\ & \angle XTU=\angle TXW={{60}^{\circ }}\text{    }\!\!\{\!\!\text{ alternate angles }\!\!\}\!\!\text{ } \\ & \angle TUQ=\angle XTU={{60}^{\circ }}\text{    }\!\!\{\!\!\text{ alternate angles }\!\!\}\!\!\text{ } \\ & \text{Consider }\vartriangle QTU \\ & \angle QTU+\angle TUQ+\angle TQU={{180}^{\circ }}\text{   }\!\!\{\!\!\text{ sum of }\angle s\text{ in a }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & {{48}^{\circ }}+{{60}^{\circ }}+\angle TQU={{180}^{\circ }} \\ & \angle TQU={{180}^{\circ }}-{{108}^{\circ }}={{72}^{\circ }} \\\end{align}$

maths year: