Question 32

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Jambmaths question: 

In the figure above, PQR is a straight line segment, PQ = QT . Triangle PQT is an isosceles triangle. $\angle QPT$is 25o . Calculate the value of $\angle RST$

Option A: 

45o

Option B: 

55o

Option C: 

25o

Option D: 

50o

Jamb Maths Solution: 

$\begin{align}  & \angle QPT=\angle QTP={{25}^{o}}\text{    }\!\!\{\!\!\text{ base angles of an issosceles triangle} \\ & \angle QPT+\angle QTP+\angle PQT={{180}^{o}}\text{   (sum of angles in a }\Delta \text{ }\!\!\}\!\!\text{ } \\ & \text{2}{{\text{5}}^{o}}+{{25}^{o}}+\angle PQT={{180}^{o}} \\ & \angle PQT={{130}^{o}} \\ & \angle PQT+\angle SQR={{180}^{o}}\text{    }\!\!\{\!\!\text{ sum of angle on straight line }\!\!\}\!\!\text{ } \\ & \text{13}{{\text{0}}^{o}}+\angle SQR={{180}^{o}} \\ & \angle SQR={{50}^{o}} \\ & \angle SQR+\angle SRQ+\angle QSR={{180}^{o}} \\ & {{50}^{o}}+{{75}^{o}}+\angle QSR={{180}^{o}} \\ & \angle QSR={{55}^{o}} \\ & \angle RST=\angle QSR={{55}^{o}} \\\end{align}$

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