Question 28

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Jambmaths question: 

The bearing of P and Q from a common point N are 020o and 300o respectively. If P and Q are also equidistance from N, find the bearing of P from Q

Option A: 

040o

Option B: 

070o

Option C: 

280o

Option D: 

320o

Jamb Maths Solution: 

$\begin{align}  & \angle QNP={{80}^{o}} \\ & \text{Since }P\text{ and }Q\text{ are equidistant from }N \\ & \left| NQ \right|=\left| NP \right| \\ & \angle NQP=\angle NPQ\text{     }\!\!\{\!\!\text{ Base angle of an issosceles triangle }\!\!\}\!\!\text{ } \\ & \text{8}{{\text{0}}^{o}}+x+x=180\text{      }\!\!\{\!\!\text{ sum of angle in }\Delta QNP\text{ }\!\!\}\!\!\text{ } \\ & 2x={{100}^{o}} \\ & x={{50}^{o}} \\ & \angle NQP=\angle NPQ=x={{50}^{o}} \\ & \angle NTQ=\angle NOQ={{60}^{o}}\text{   }\!\!\{\!\!\text{ alternate angles }\!\!\}\!\!\text{ } \\ & \text{The bearing of }P\text{ from }Q={{180}^{o}}-({{50}^{o}}+{{60}^{o}})={{070}^{\circ }}\text{   } \\ & \text{ }\!\!\{\!\!\text{ sum of angles on straight line }\!\!\}\!\!\text{ } \\\end{align}$

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