Question 6:
In Diagram below, PQRST is a straight line. Find the value of x.

Solution:
$\begin{array}{l}\text{Interior angle of }R=\left({180}^o-{76}^o\right)÷2={52}^o\\ \angle x=\text{exterior angle of }R\left(\text{corresponding angles}\right)\\ \text{Hence, }x={76}^o+{52}^o={128}^o\end{array}$

Question 7:
In Diagram below, find the value of y.



Solution:


$\begin{array}{l}\angle ABC=\angle BCD\\ ={108}^o\text{ (alternate angle)}\\ y^o+{130}^o+{108}^o={360}^o\\ y^o={360}^o-{238}^o\\ y^o={122}^o\end{array}$

Question 8:
In Diagram below, PSR and QST are straight lines.
Find the value of x.

Solution:
$\begin{array}{l}\angle UST+\angle STV={180}^o\\ \angle UST={180}^o-{116}^o={64}^o\\ \angle PST=\angle QSR\\ x^o+\angle UST={135}^o\\ x^o+\angle UST={135}^o\\ x^o={71}^o\\ x=71\end{array}$

Question 9:
In Diagram below, PWV is a straight line.

(a) Which line is perpendicular to line PWV?
(b) State the value of  ∠ RWU.

Solution:
(a) SW

(b) ∠ RWU = 13^o + 29^o + 20^o = 62^o

 

Question 10:
In Diagram below, UVW is a straight line.


(a) Which line is parallel to line TU?
(b) State the value of  ∠ QVS.

Solution:
(a) QV

(b) ∠ QVS = 8^o + 18^o = 26^o

 

In Diagram below, PQRS, ABQC and KRL are straight lines.
$\begin{array}{l}\angle KRB+\angle ABR={180}^o\leftarrow \boxed{\text{Sum of interior angles}}\\ \angle KRB+{105}^o={180}^o\\ \angle KRB={180}^o-{105}^o\\ \angle KRB={75}^o\\ \angle BRQ={40}^o\leftarrow \boxed{\text{Corresponding angle}}\\ x^o=\angle SRL=\angle KRQ\leftarrow \boxed{\begin{array}{l}\text{Vertically}\\ \text{opposite angle}\end{array}}\\ x^o={75}^o+{40}^o\\ ={115}^o\\ x=115\end{array}$

(a) In Diagram below, PQR and SQT are straight lines.
$\begin{array}{l}\angle QRT={57}^o\\ x^o+2x^o={180}^o-{57}^o\\ \text{3}{x}^o={123}^o\\ x^o=\frac{{123}^o}{3}\\ x^o={41}^o\\ x=41\end{array}$

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

Find the values of x and y.

(C) Alternate angles

Corresponding angles are easily identified by the pattern “F” as shown.

$$\begin{array}{l}\angle WSQ={180}^o-{150}^o\\ ={30}^o\leftarrow \boxed{\text{Supplementary angle}}\\ \angle XTU=\angle WSQ+\angle x\leftarrow \boxed{\text{Corresponding angle}}\\ {75}^o={30}^o+\angle x\\ \angle x={75}^o-{30}^o\\ \angle x={45}^o\\ x=45\end{array}$$