Study P5 Mathematics Angle Properties - Geniebook

# Angle Properties

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

In this article, the learning objectives are:

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 $$180º$$.

$$∠a + ∠b = 180º$$     (angles on a straight line)

Question 1:

AB is a straight line. Find $$∠b$$.

Solution:

$$∠b + 45º = 180º$$     (angles on a straight line)

\begin{align*} ∠b &= 180º - 45º\\ &= 135º\\ \end{align*}

$$135º$$

Question 2:

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

Solution:

$$∠y + 88º + 36º = 180º$$     (angles on a straight line)

$$∠y = 180º - 88º - 36º$$

$$= 56º$$

$$56º$$

Question 3:

In the figure below, $$KLM$$ is a straight line. $$∠KLN = 122º$$ and $$∠JLM = 105º$$. Find $$∠JLN$$

Solution:

\begin{align*} ∠KLN + ∠NLM &= 180º​​​​​​​ (angles \;on \;a \;straight \;line)\\ 122º + ∠NLM &= 180º \\ ∠NLM &= 180º - 122º \\ ∠NLM &= 58º\\ \end{align*}

\begin{align*} ∠JLN + ∠NLM &= 105º\\ ∠JLN + 58º &= 105º\\ ​​​​​​​∠JLN &= 105º - 58º \\ ∠JLN &= 47º\\ \end{align*}

$$47º$$

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

In this article, we learnt about the different types of angles and their properties.

• 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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