Question 22

loading...

Prev/Next links

waecmaths question: 

In the diagram, TS is a tangent to the circle at S . $\left| PR \right|=\left| RS \right|$ and $\angle PQR={{117}^{\circ }}$. Calculate $\angle PST$

Option A: 

54o

Option B: 

44o

Option C: 

34o

Option D: 

27o

waecmaths solution: 

$\begin{align}  & \angle PSR={{180}^{\circ }}-\angle PQR\text{  }\!\!\{\!\!\text{ sum of opp}\text{. }\angle s\text{ in cyclic quad }\!\!\}\!\!\text{ } \\ & \angle PSR={{180}^{\circ }}-{{117}^{\circ }}={{63}^{\circ }} \\ & \angle RPS=\angle PSR={{63}^{\circ }}\text{    }\!\!\{\!\!\text{ Base }\angle s\text{ of Isso}\text{. }\vartriangle \} \\ & \angle RPS+\angle PSR+\angle PRS={{180}^{\circ }}^{\circ }\text{   }\!\!\{\!\!\text{ sum of }\angle s\text{ in }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & \text{6}{{\text{3}}^{\circ }}+{{63}^{\circ }}+\angle PRS={{180}^{\circ }} \\ & \angle PRS={{180}^{\circ }}-{{126}^{\circ }}={{54}^{\circ }} \\ & \angle PST=\angle PRS={{54}^{\circ }}\text{   }\!\!\{\!\!\text{ Angles in opposite segment }\!\!\}\!\!\text{ } \\\end{align}$

maths year: 
maths topics: