Chapter 7.3, Problem 5PSA

Solution

a.

The measure of $\angle JKP$

The measure of $\angle JKP$ is ${70}^{\circ }$

Given: K is mid-point, P is a mid-point, $m\angle M=70$ , $m\angle JKP=y+15$ and $m\angle JPK=y-10$ is provided in the question which will help to find the solution.

Concept used: Here, we are using the concept of corresponding angles.

Calculation: Here, JK=KM ( $\because$ K is a mid-point)

And, JP=PQ ( $\because$ P is a mid-point)

Here, we know that $\angle JKP=\angle M$ (corresponding angles)

So, $m\angle JKP={70}^{\circ }$

Conclusion: Hence, the measure of $\angle JKP$ is ${70}^{\circ }$

b.

The measure of $m\angle JPK$

The measure of $m\angle JPK$ is ${45}^{\circ }$

Given: K is mid-point, P is a mid-point, $m\angle M=70$ , $m\angle JKP=y+15$ and $m\angle JPK=y-10$ is provided in the question which will help to find the solution.

Calculation: Here, $\angle JKP=y+15$ (given)

And, $\angle JKP$ is ${70}^{\circ }$ (found above)

So, ${70}^{\circ }=y+15$

  $y={55}^{\circ }$

We know that, $m\angle JPK=y-10$ (given)

➔ $m\angle JPK={55}^{\circ }-10$

➔ $m\angle JPK={45}^{\circ }$

Conclusion: Hence, the measure of $m\angle JPK$ is ${45}^{\circ }$

c.

The measure of $\angle J$

The measure of $\angle J$ is ${65}^{\circ }$

Given: K is mid-point, P is a mid-point, $m\angle M=70$ , $m\angle JKP=y+15$ and $m\angle JPK=y-10$ is provided in the question which will help to find the solution.

Concept used: Here, we are using the concept of angle sum property.

Calculation: Here, $m\angle JPK=m\angle D={45}^{\circ }$ (corresponding angles)

And, $\angle J+\angle M+\angle D={180}^{\circ }$ (angle sum property)

➔ $\angle J+{70}^{\circ }+{45}^{\circ }={180}^{\circ }$

➔ $\angle J={180}^{\circ }-{115}^{\circ }$

➔ $m\angle J={65}^{\circ }$

Conclusion: Hence, the measure of $\angle J$ is ${65}^{\circ }$