Question 26


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Jambmaths question: 

In the figure above $TS\parallel XY$and XY = TY, $\angle STZ={{34}^{\circ }},\text{ }\angle TXY={{47}^{\circ }}$find the angle marked n

Jamb Maths Solution: 

$\begin{align}  & \angle TXY=\angle XTY={{47}^{\circ }}\text{   (Base }\angle \text{s of issocele }\vartriangle \text{)} \\ & \angle XZT=\angle STZ={{34}^{\circ }}\text{    (Alternate angles)} \\ & \angle TXY+\angle XZT+\angle XTZ={{180}^{\circ }}\text{  (Sum of angles in a }\vartriangle \text{)} \\ & {{47}^{\circ }}+{{34}^{\circ }}+\angle XTZ={{180}^{\circ }} \\ & \angle XTZ={{180}^{\circ }}-({{47}^{\circ }}+{{34}^{\circ }})={{99}^{\circ }} \\ & \therefore \angle XTY+\angle YTZ={{99}^{\circ }} \\ & \angle YTZ={{99}^{\circ }} \\ & \angle YTZ={{99}^{\circ }}-{{47}^{\circ }}={{52}^{\circ }}\text{ } \\\end{align}$

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