Angle Properties

In this article, we are going to study about Angles as per the Primary 5 Maths requirements. We will be learning about the angle properties involving lines.

1. Adjacent angles on a straight line
2. Angles at a point
3. Vertically opposite angles

1. Adjacent Angles On A Straight Line

Adjacent angles on a straight line add up to $180^\circ$.

\begin{align} \mathrm{\angle{a} + \angle{b}} = \mathrm{180^\circ} && \text{(angles on a straight line)} \end{align}

Question 1:

$\mathrm{AB}$ is a straight line. Find $\mathrm{\angle{b}}$.

Solution:

\begin{align*} \mathrm{\angle{b} + 45^\circ} &= \mathrm{180^\circ} & \text{(angles on a straight line)} \\[2ex] \mathrm{\angle{b}} &= \mathrm{180^\circ - 45^\circ}\\[2ex] &= \mathrm{135^\circ} \end{align*}

$135^\circ$

Question 2:

The following figure is not drawn to scale. Find $\mathrm{\angle{y}}$.

Solution:

\begin{align} \mathrm{\angle{y} + 88^\circ + 36^\circ} &= \mathrm{180^\circ} & \text{(angles on a straight line)} \\[2ex] \mathrm{\angle y} &= \mathrm{180^\circ - 88^\circ - 36^\circ} \\[2ex] &= \mathrm{56 ^\circ} \end{align}

$\mathrm{56^\circ}$

Question 3:

In the figure below, $\mathrm{KLM}$ is a straight line. $\mathrm{\angle{KLN}} = \mathrm{122^\circ}$ and $\mathrm{\angle{JLM}} = \mathrm{105^\circ}$. Find $\mathrm{\angle{JLN}}$

Solution:

\begin{align*} \mathrm{\angle{KLN} + \angle{NLM}} &= \mathrm{180^\circ}​​​​​​​ & \text{(angles on a straight line)}\\[2ex] \mathrm{122^\circ + \angle{NLM}} &= \mathrm{180^\circ} \\[2ex] \mathrm{\angle{NLM}} &= \mathrm{180^\circ - 122^\circ} \\[2ex] \mathrm{\angle{NLM}} &= \mathrm{58^\circ} \end{align*}

\begin{align*} \mathrm{\angle{JLN} + \angle{NLM}} &= \mathrm{105^\circ}\\[2ex] \mathrm{\angle{JLN} + 58^\circ} &= \mathrm{105^\circ}\\[2ex] \mathrm{​​​​​​​\angle{JLN}} &= \mathrm{105^\circ - 58^\circ}\\[2ex] \mathrm{\angle{JLN}} &= \mathrm{47^\circ} \end{align*}

$\mathrm{47^\circ}$

OR

\begin{align*} ∠JLM + ∠JLK &= 180º \;\;\;\;\;(angles \;on \;a \;straight \;line)\\ 105º + ∠JLK &= 180º\\ ∠JLK &= 180º - 105º\\ ∠JLK &= 75º\\ \end{align*}

\begin{align*} ∠JLK + ∠JLN &= 122º\\ 75º + ∠JLN &= 122º\\ ∠JLN &= 122º − 75º \\ &= 47º\\ \end {align*}

$47º$

2. Angles At A Point

Angles at a point add to $360º$

$∠a + ∠b + ∠c = 360º$     (angles at a point)

Question 1:

Find $∠c$.

Solution:

\begin{align*} 140º + 60º + ∠c &= 360º \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(angles \;at \;a \;point)\\ ∠c &= 360º - 140º - 60º \\ &= 160º \\ \end{align*}

$160º$

Question 2:

Find the sum of $∠d$ and $∠e$.

Solution:

\begin{align*} 75º + 75º + ∠d + ∠e &= 360º \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(angles \;at \;a \;point)\\ ∠d + ∠e &= 360º - 75º - 75º \\ &= 210º \end{align*}

$210º$

Question 3:

The following figure is not drawn to scale. Find $∠z$

Solution:

\begin{align*} ∠z + 86º + 81º + 90º &= 360º \;\;\;\;\;\;\;\;\;(angles \;at \;a \;point) \\ ​​​​​​​∠z &= 360º - 90º - 86º - 81º \\ ∠z &= 103º \\ \end{align*}

$103º$

3. Vertically Opposite Angles

When two straight lines intersect, the opposite angles are equal. The point where they meet is called the vertex.

$∠a = ∠c$     (vertically opposite angles)

$∠b = ∠d$     (vertically opposite angles)

Question 1:

$AB$ and $CD$ are straight lines. Name the 2 pairs of angles that are equal.

Solution:

$∠a$ and $∠c$ are vertically opposite angles.

$∠b$ and $∠d$ are also vertically opposite angles.

As per the properties of vertically opposite angles,

$∠a = ∠c$ and $∠b = ∠d$

$∠a = ∠c$ and $∠b = ∠d$

Question 2:

$AB$ and $CD$ are straight lines. Find $∠b$.

Solution:

$AB$ and $CD$ are two straight lines that meet.

$∠b = 35º$     (vertically opposite angles)

35º

Question 3:

In the figure below, $AB$ and $CD$ are straight lines. $∠AOC = 25º$

1. Find $∠BOE$.
2. Find $∠AOD$.

Solution:

(a)

$AB$ is a straight line.

\begin{align*} ∠AOC + ∠COE + ∠BOE &= 180º \;\;\;\;\;\;\;\;\;(angles \;on \;a \;straight \;line)\\ ∠BOE &= 180º - 25º - 90º\\ &= 65º\\ \end{align*}

(b)

$AB$ and $CD$ are two straight lines that intersect.

\begin{align*} ∠AOD &= ∠BOC \;\;\;\;\;\;\;\;\;(vertically \;opposite \;angles) \\ ∠AOD &= 90º + 25º \\ &= 115º \\ \end{align*}

(a) $65º$

(b) $115º$

Question 4:

The figure below is not drawn to scale. $AD$ and $BC$ are straight lines. The ratio of $∠a$ to $∠b$ is $3 : 2$. Find the difference between $∠a$ and $∠c$

Solution:

$BC$ is a straight line.

\begin{align*} ∠c + ∠AOC &= 180º \;\;\;\;\;\;\;(angles \;on \;a \;straight \;line)\\ ∠c &= 180º - 155º \\ &= 25º \\ \end{align*}

$AD$ and $BC$ are two straight lines that meet.

\begin{align*} ∠a + ∠b &= 155º \;\;\;\;\;\;\;(vertically \;opposite \;angles)\\\\ ∠a : ∠b &= 3 : 2 \\ \\ 5 \;units &= 155º \\ 1 \;unit &= 155º \div5 \\ &= 31º \\ \\ ∠a &= 3 \;units \\ &= 3 × 31º \\ &= 93º \\ \\ ∠b &= 2 \;units \\ &= 2 × 31º \\ &= 62º \\ \end{align*}

Difference between $∠a$ and $∠c$

\begin{align*} &= 93º - 25º \\ &= 68º \\ \end{align*}

$68º$

Conclusion

• Properties of angles involving lines
 Angles On A Straight Line Angles At A Point Vertically Opposite Angles $∠a + ∠b = 180º$ $∠a + ∠b + ∠c = 360º$ $∠a = ∠c\\ ∠b = ∠d$

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