Jambmaths question:
In the diagram above, calculate the value of x
Option A:
120o
Option B:
60o
Option C:
140o
Option D:
100o
Jamb Maths Solution:
$\begin{align} & \angle EFB+\angle BFG={{180}^{\circ }}\text{ }\!\!\{\!\!\text{ sum of angles on a straight line }\!\!\}\!\!\text{ } \\ & \text{10}{{\text{0}}^{\circ }}+\angle BFG={{180}^{\circ }} \\ & \angle BFG={{80}^{\circ }} \\ & \angle FBG+\angle BFG=\angle BGH\text{ }\!\!\{\!\!\text{ Sum of two opp}\text{. angles of a }\vartriangle \} \\ & {{80}^{\circ }}+{{40}^{\circ }}=\angle BGH \\ & \angle BGH={{120}^{\circ }} \\ & x=\angle BCD=\angle BGH={{120}^{\circ }} \\\end{align}$
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