The Solution:
Given the figure below:
Solving for arc JK:
By angle subtends at the center of a circle is twice that subtends on the circumference, we have that:
[tex]2\times118=236^o[/tex]
Subtracting 236 from 360, we get
[tex]arcJK=360-236=124^o\text{ (angle at a point)}[/tex]
So, the correct answer for question 7 is [option 4]
To answer Question 8:
[tex]\begin{gathered} \angle JKM=\frac{70}{2}=35^o\text{ } \\ \\ \text{ Reason: Angle subtends at the center is twice that subtends} \\ \text{ at the circumference of the circle.} \end{gathered}[/tex]
Similarly,
[tex]\begin{gathered} \angle KJL=\frac{60}{2}=30^o \\ \\ \text{Reason: Angle subtends at the center is twice that subtends} \\ \text{ at the circumference of the circle.} \end{gathered}[/tex][tex]\text{ To get }\angle1[/tex][tex]\begin{gathered} \angle1=180-(\angle JLM+\angle KML) \\ \text{ Reason: sum of angles in a triangle.} \end{gathered}[/tex]
[tex]\begin{gathered} \angle JLM=\angle JKM=35^o \\ \text{ Reason: angles on the same segments.} \\ \text{ Similarly,} \\ \angle KML=\angle KJL=30^o \\ \text{ Reason: angles on the same segments.} \end{gathered}[/tex]
[tex]\begin{gathered} \angle1=180-(35+30) \\ \text{ Sum of angles in a triangle.} \\ \angle1=180-65=115^o \end{gathered}[/tex]
Therefore, the correct answer to Question 8 is 115 degrees.
