Question 27

loading...

Prev/Next links

Jambmaths question: 

In the diagram above, PST is a straight line $PQ=QS=RS$, if $\angle RST={{72}^{\circ }}$, find x 

Option A: 

36o

Option B: 

18o

Option C: 

72o

Option D: 

24o

Jamb Maths Solution: 

$\begin{align}  & \text{Consider }\triangle QPS \\ & \angle QPS=\angle QSP=x\text{     }\!\!\{\!\!\text{ }Base\text{ }angles\text{ }of\text{ }an\text{ }issoscele\text{ }\vartriangle \} \\ & \angle RQS=\angle QPS+\angle QSP=2x\text{  }\{Sum\text{ }of\text{ }the\text{ }opp.int.\text{ }\angle =opp.ext.\text{ }\angle \} \\ & \angle RQS=\angle SRQ=2x\text{   }\{Base\text{ }angle\text{ }of\text{ }an\text{ }isosceles\vartriangle \text{ }\} \\ & \text{Consider }\vartriangle RQS \\ & \angle RQS+\angle SRQ+\angle RSQ={{180}^{o}}\text{  }\!\!\{\!\!\text{ sum of angles in }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & 2{{x}^{o}}+2{{x}^{o}}+\angle RSQ={{180}^{0}} \\ & \angle RSQ={{180}^{o}}-4{{x}^{o}}\text{Consider line }TSP \\ & {{72}^{o}}+\angle RSQ+\angle QSP={{180}^{0}}\text{  }\!\!\{\!\!\text{ angles on a straight line }\!\!\}\!\!\text{ } \\ & \text{7}{{\text{2}}^{o}}+x+180-4{{x}^{o}}={{180}^{o}} \\ & 3{{x}^{o}}={{72}^{o}},\text{  }x={{24}^{o}} \\\end{align}$

Year of Exam: