**1.2.1 Angles and Lines II, PT3 Practice**

**Question 1:**

*PQRS*,

*ABQC*and

*KRL*are straight lines.

Find the value of

$\begin{array}{l}\angle KRB+\angle ABR={180}^{o}\leftarrow \overline{)\text{Sum of interior angles}}\\ \angle KRB+{105}^{o}={180}^{o}\\ \text{}\angle KRB={180}^{o}-{105}^{o}\\ \text{}\angle KRB={75}^{o}\\ \angle BRQ={40}^{o}\leftarrow \overline{)\text{Corresponding angle}}\\ {x}^{o}=\angle SRL=\angle KRQ\leftarrow \overline{)\begin{array}{l}\text{Vertically}\\ \text{opposite angle}\end{array}}\\ {x}^{o}={75}^{o}+{40}^{o}\\ \text{}={115}^{o}\\ \text{}x=115\end{array}$
*x*.**Question 2:**

*PQR*and

*SQT*are straight lines.

Find the value of

*x*.(b) In Diagram below,

*PQRS*is a straight line. Find the value of

*x*.

*Solution:***(a)**

$\begin{array}{l}\angle PQT=\angle RQS\\ {x}^{o}+{90}^{o}={155}^{o}\\ \text{}{x}^{o}={155}^{o}-{90}^{o}\\ \text{}={65}^{o}\\ \text{}x=65\end{array}$

(b)

(b)

**Question 3:**

$\begin{array}{l}{x}^{o}=\angle QRT+\angle TRS\\ \angle QRT={40}^{o}\\ \angle TRS={180}^{o}-{135}^{o}\\ \text{}={45}^{o}\\ \text{Hence,}{x}^{o}={40}^{o}+{45}^{o}\\ \text{}{x}^{o}={85}^{o}\\ \text{}x=85\end{array}$

**Question 4:**

*Solution:*40

^{o}+ 80^{o}+*x*^{o }+*x*^{o}= 180^{o}2

*x*^{o}= 180^{o}– 120^{o }2

*x*^{o}= 60^{o}

*x*

^{o }= 30

^{o}

*x*

**= 30**

**Question 5:**

*x*and

*y.*

Solution:Solution:

$\begin{array}{l}\angle PRQ={180}^{o}-{110}^{o}={70}^{o}\\ \angle PRQ=\angle PQR={70}^{o}\\ \text{}{y}^{o}={180}^{o}-{70}^{o}-{70}^{o}\\ \text{}={40}^{o}\\ \text{}y=40\\ {x}^{o}={110}^{o}-{40}^{o}\\ \text{}={70}^{o}\\ \text{}x=70\end{array}$