# 1.2.1 Angles and Lines II, PT3 Practice

 
 
1.2.1 Angles and Lines II, PT3 Practice
 

 
Question 1:
 In Diagram below, PQRS, ABQC and KRL are straight lines.
 
 
 
 
 
Find the value of x.
 
 

  Solution:

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

 
 
 
 
Question 2:
 (a) In Diagram below, 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)
    $\begin{array}{l}\angle QRT={57}^{o}\\ {x}^{o}+2{x}^{o}={180}^{o}-{57}^{o}\\ \text{3}{x}^{o}={123}^{o}\\ \text{}{x}^{o}=\frac{{123}^{o}}{3}\\ \text{}{x}^{o}={41}^{o}\\ \text{}x=41\end{array}$
 

 
 
 
 
Question 3:
 
In Diagram below, PQR is a straight line. Find the value of x.
 
 Solution:
 
 
 
 
 
 
 
    $\begin{array}{l}{x}^{o}=\angle QRT+\angle TRS\\ \angle QRT={40}^{o}\\ \angle TRS={180}^{o}-{135}^{o}\\ \text{}={45}^{o}\\ \text{Hence,}{x}^{o}={40}^{o}+{45}^{o}\\ \text{}{x}^{o}={85}^{o}\\ \text{}x=85\end{array}$  
 
 
 
 
 
Question 4:
 
In Diagram below, find the value of x.
 
 
 
 
 
 
 
Solution:
 
40o + 80o + xo + xo = 180o
 
2xo = 180o – 120o 
 
2xo = 60o
 
x= 30o
 
x = 30
 
 
 
 
 
 
Question 5:
 
In Diagram below, PQR is an isosceles triangle and QRS is a straight line.
 
 
 
 Find the values of x and y.
 
 

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

 

 

1. NICE!!!!

2. tq..