Question

Please find arc JK and angle 1. Ignore the entered answers.

Answer 1

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:

`2*118=236^o`

Subtracting 236 from 360, we get

`arcJK=360-236=124^o\text{ (angle at a point)}`

So, the correct answer for question 7 is [option 4]

To answer Question 8:

`\begin{gathered} \angle JKM=(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}`

Similarly,

`\begin{gathered} \angle KJL=(60)/(2)=30^o \\ \\ \text{Reason: Angle subtends at the center is twice that subtends} \\ \text{ at the circumference of the circle.} \end{gathered}`
`\text{ To get }\angle1`
`\begin{gathered} \angle1=180-(\angle JLM+\angle KML) \\ \text{ Reason: sum of angles in a triangle.} \end{gathered}`

`\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}`

`\begin{gathered} \angle1=180-(35+30) \\ \text{ Sum of angles in a triangle.} \\ \angle1=180-65=115^o \end{gathered}`

Therefore, the correct answer to Question 8 is 115 degrees.