Question 24

loading...

Prev/Next links

Jambmaths question: 

In the diagram above, find the value of x

Option A: 

15o

Option B: 

30o

Option C: 

40o

Option D: 

45o

Jamb Maths Solution: 

$\begin{align}  & \text{Consider }\Delta CDE \\ & \angle CED=\angle CDE={{15}^{\circ }}\text{ }\!\!\{\!\!\text{ Base angle of Isosceles }\Delta \text{ }\!\!\}\!\!\text{ } \\ & \angle ECD={{180}^{\circ }}-\angle CED-\angle CDE\text{  }\!\!\{\!\!\text{ sum of angles in }\Delta \} \\ & \angle ECD={{180}^{\circ }}-{{15}^{\circ }}-{{15}^{\circ }}={{150}^{\circ }} \\ & \angle ECD+\angle BCA={{180}^{\circ }}\text{ }\!\!\{\!\!\text{ Sum of angles on a straight line }\!\!\}\!\!\text{ } \\ & \text{15}{{\text{0}}^{\circ }}+\angle BCA={{180}^{\circ }} \\ & \angle BCA={{30}^{\circ }} \\ & \text{Consider }\Delta ABC \\ & \angle BAC+\angle ABC+\angle BCA={{180}^{\circ }} \\ & x+{{110}^{\circ }}+{{30}^{\circ }}={{180}^{\circ }} \\ & x={{40}^{\circ }} \\\end{align}$

Year of Exam: