Angles on a Straight Line and at a Point

Adjacent angles on a straight line add up to 180° and the angles at a point add up to 360°. Use these two facts, with the reason written in every step, to find unknown angles.

Before you start

An angle is named with three letters, the vertex in the middle: $\angle AOB$ is the angle at $O$ between the arms $OA$ and $OB$ (Lesson 11.1).
A straight angle, a half turn, is $180^\circ$ , and a complete turn is $360^\circ$ . A reflex angle is greater than $180^\circ$ and less than $360^\circ$ .
Two angles that add up to $180^\circ$ are supplementary.
An equation is solved by doing the same to both sides (Chapter 9): from $5x + 30 = 180$ , subtract $30$ from both sides to get $5x = 150$ , then divide both sides by $5$ to get $x = 30$ .

Angles on a Straight Line

Two angles are adjacent when they have the same vertex and share one arm, and lie on opposite sides of that arm, so that they do not overlap. In the figure below, $\angle AOC$ and $\angle COD$ are adjacent: they share the vertex $O$ and the arm $OC$ .

The straight line $AOB$ makes a straight angle at $O$ : a half turn, $180^\circ$ . The arms $OC$ and $OD$ cut that half turn into three angles side by side, each adjacent to the next, with no gap between them and no overlap. Together they still make the half turn, so they add up to $180^\circ$ .

As in Lesson 11.1, a letter written in an angle with the degree sign, such as $a^\circ$ or $x^\circ$ , stands for a number: if $a = 65$ , the angle is $65^\circ$ .

Angles on a straight line

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

In the figure, $AOB$ is a straight line, so $a + b + c = 180$ .

In working, the reason is written in brackets as (adj. ∠s on a str. line), short for "adjacent angles on a straight line".

Writing your working

Write each step on its own line. After a step that uses an angle fact, write the fact as a reason in brackets. Solve the equation as in Chapter 9, then check that the angles you have found add up to the right total.

The equation is about the numbers of degrees, so it is written without degree signs: $x + 110 = 180$ . The check is about the angles themselves, so it has them: $70^\circ + 110^\circ = 180^\circ$ .

Example 1: One unknown angle

In the figure, $AOB$ is a straight line. Find $x$ .

Solution

$\angle AOC$ and $\angle COB$ are adjacent angles on the straight line $AOB$ .

$x + 110 = 180$ (adj. ∠s on a str. line)

$x = 180 - 110 = 70$

Check: $70^\circ + 110^\circ = 180^\circ$ .

$x = 70$ : the unknown angle $\angle COB$ is $70^\circ$ .

Example 2: Three angles on a straight line

In the figure, $AOB$ is a straight line. Find $y$ .

Solution

$60 + 45 + y = 180$ (adj. ∠s on a str. line)

Add the known angles: $105 + y = 180$

$y = 180 - 105 = 75$

Check: $60^\circ + 45^\circ + 75^\circ = 180^\circ$ .

$y = 75$ : $\angle DOB = 75^\circ$ .

When the same letter is written in two or more angles of a figure, those angles are equal. Count how many of them there are.

Example 3: Equal unknown angles

In the figure, $AOB$ is a straight line. Find $a$ .

Solution

There are three angles marked $a^\circ$ , so together they are $3a^\circ$ .

$54 + a + a + a = 180$ (adj. ∠s on a str. line)

$3a + 54 = 180$

Subtract $54$ from both sides: $3a = 126$

Divide both sides by $3$ : $a = 42$

Check: $54^\circ + 42^\circ + 42^\circ + 42^\circ = 180^\circ$ .

$a = 42$ : each of the three equal angles is $42^\circ$ .

Adjacent, but not on a straight line

The rule is about adjacent angles whose outer arms make a straight line. In the figure below, $\angle COB$ is above the line $AB$ and $\angle BOD$ is below it. They are adjacent (they share the vertex $O$ and the arm $OB$ ), but their outer arms $OC$ and $OD$ do not make a straight line, so the rule says nothing about them. Here $60^\circ + 130^\circ = 190^\circ$ , not $180^\circ$ .

Forming and Solving an Equation

Often the angles are given as expressions in $x$ , such as $2x^\circ$ or $(3x - 10)^\circ$ . The rule still gives an equation: write the angles added together equal to $180$ , collect the like terms, and solve it. Then substitute $x$ back to find each angle, and check that the angles add up to $180^\circ$ .

Example 4: Two angles in terms of $x$

In the figure, $AOB$ is a straight line. Find $x$ , and the size of each angle.

Solution

$2x + 3x = 180$ (adj. ∠s on a str. line)

Collect like terms: $5x = 180$

Divide both sides by $5$ : $x = 36$

$\angle AOC = 2x^\circ = 2 \times 36^\circ = 72^\circ$ and $\angle COB = 3x^\circ = 3 \times 36^\circ = 108^\circ$ .

Check: $72^\circ + 108^\circ = 180^\circ$ .

$x = 36$ ; $\angle AOC = 72^\circ$ and $\angle COB = 108^\circ$ .

Example 5: Expressions with brackets

In the figure, $AOB$ is a straight line. Find $x$ , and the size of each angle.

Solution

$(3x - 10) + 40 + (x + 30) = 180$ (adj. ∠s on a str. line)

Remove the brackets: $3x - 10 + 40 + x + 30 = 180$

Collect like terms ( $3x + x = 4x$ and $-10 + 40 + 30 = 60$ ): $4x + 60 = 180$

Subtract $60$ from both sides: $4x = 120$

Divide both sides by $4$ : $x = 30$

$\angle AOC = (3 \times 30 - 10)^\circ = 80^\circ$ , $\angle COD = 40^\circ$ and $\angle DOB = (30 + 30)^\circ = 60^\circ$ .

Check: $80^\circ + 40^\circ + 60^\circ = 180^\circ$ .

$x = 30$ ; $\angle AOC = 80^\circ$ , $\angle COD = 40^\circ$ and $\angle DOB = 60^\circ$ .

Is It a Straight Line?

The rule also works the other way round. The angles, not your eyes, show whether $A$ , $O$ and $B$ lie on a straight line, however straight $AOB$ looks: a bend of one degree can hardly be seen.

Testing for a straight line

If $\angle AOC + \angle COB = 180^\circ$ , where the two angles are adjacent, then $AOB$ is a straight line.

If $\angle AOC + \angle COB$ is not $180^\circ$ , then $AOB$ is not a straight line.

Example 6: Do the points lie on a straight line?

(a) In the first figure, $\angle AOC = 48^\circ$ and $\angle COB = 132^\circ$ . Is $AOB$ a straight line?

(b) In the second figure, $\angle POR = 48^\circ$ and $\angle ROQ = 133^\circ$ . Is $POQ$ a straight line?

Solution

(a) $\angle AOC$ and $\angle COB$ are adjacent. $48^\circ + 132^\circ = 180^\circ$ .

They add up to exactly $180^\circ$ , a half turn, so $OA$ and $OB$ point in opposite directions.

(b) $\angle POR$ and $\angle ROQ$ are adjacent. $48^\circ + 133^\circ = 181^\circ$ .

They add up to $181^\circ$ , not $180^\circ$ , so $POQ$ bends by $1^\circ$ at $O$ . It looks straight, but it is not.

(a) Yes: $\angle AOC + \angle COB = 180^\circ$ , so $AOB$ is a straight line. (b) No: $\angle POR + \angle ROQ = 181^\circ$ , not $180^\circ$ , so $POQ$ is not a straight line.

Angles at a Point

Now go all the way round. A complete turn about a point is $360^\circ$ . When arms drawn from a point $O$ cut the complete turn into angles, side by side all the way round, with no gap between them and no overlap, the angles together still make the complete turn. So they add up to $360^\circ$ . These are the angles at a point.

Angles at a point

Angles at a point, which together make one complete turn about the point, add up to $360^\circ$ .

In the figure, $a + b + c + d = 360$ .

In working, the reason is written in brackets as (∠s at a pt), short for "angles at a point".

When the rule does not apply

The angles must go all the way round. In the first figure below, $\angle AOB = 30^\circ$ and $\angle BOC = 40^\circ$ meet at $O$ , but they leave a gap: reflex $\angle AOC$ , on the other side. Only with that angle do they make the complete turn: $30^\circ + 40^\circ + 290^\circ = 360^\circ$ . (In the same way, the two angles of Example 1 meet at $O$ and add up to only $180^\circ$ : the other half turn lies below the line.)

The angles must not overlap. In the second figure, $\angle AOC$ (the large outer arc) is made up of $\angle AOB$ and $\angle BOC$ , so $\angle AOC = 50^\circ + 80^\circ = 130^\circ$ . Adding $\angle AOB + \angle BOC + \angle AOC$ counts the same turn twice, and it gives $260^\circ$ , which means nothing.

Example 7: One unknown angle at a point

In the figure, four arms meet at the point $O$ . Find $x$ .

Solution

The small square marks a right angle: $\angle AOB = 90^\circ$ .

The four angles go all the way round $O$ .

$90 + 120 + 80 + x = 360$ (∠s at a pt)

Add the known angles: $290 + x = 360$

$x = 360 - 290 = 70$

Check: $90^\circ + 120^\circ + 80^\circ + 70^\circ = 360^\circ$ .

$x = 70$ : $\angle DOA = 70^\circ$ .

Example 8: Angles at a point in terms of $x$

In the figure, four arms meet at the point $O$ . Find $x$ , and the size of each angle.

Solution

$(x + 25) + 2x + (3x - 15) + 110 = 360$ (∠s at a pt)

Remove the brackets: $x + 25 + 2x + 3x - 15 + 110 = 360$

Collect like terms ( $x + 2x + 3x = 6x$ and $25 - 15 + 110 = 120$ ): $6x + 120 = 360$

Subtract $120$ from both sides: $6x = 240$

Divide both sides by $6$ : $x = 40$

$\angle AOB = (40 + 25)^\circ = 65^\circ$ , $\angle BOC = 2 \times 40^\circ = 80^\circ$ , $\angle COD = (3 \times 40 - 15)^\circ = 105^\circ$ and $\angle DOA = 110^\circ$ .

Check: $65^\circ + 80^\circ + 105^\circ + 110^\circ = 360^\circ$ .

$x = 40$ ; the four angles are $65^\circ$ , $80^\circ$ , $105^\circ$ and $110^\circ$ .

When angles are in a given ratio, such as $4 : 5 : 6$ , write them as $4x^\circ$ , $5x^\circ$ and $6x^\circ$ : they then have the ratio $4 : 5 : 6$ whatever $x$ is. The rule gives the equation that finds $x$ .

Example 9: Angles in a given ratio

In the figure, three arms meet at the point $O$ , and $\angle AOB : \angle BOC : \angle COA = 4 : 5 : 6$ . Find the three angles.

Solution

Let $\angle AOB = 4x^\circ$ , $\angle BOC = 5x^\circ$ and $\angle COA = 6x^\circ$ .

$4x + 5x + 6x = 360$ (∠s at a pt)

Collect like terms: $15x = 360$

Divide both sides by $15$ : $x = 24$

$\angle AOB = 4 \times 24^\circ = 96^\circ$ , $\angle BOC = 5 \times 24^\circ = 120^\circ$ and $\angle COA = 6 \times 24^\circ = 144^\circ$ .

Check: $96^\circ + 120^\circ + 144^\circ = 360^\circ$ , and dividing each by $24$ gives back $4 : 5 : 6$ .

$\angle AOB = 96^\circ$ , $\angle BOC = 120^\circ$ and $\angle COA = 144^\circ$ .

One of the angles at a point may be a reflex angle, the large angle on the outside. An angle and the reflex angle on its other side together make the complete turn, so reflex $\angle AOB = 360^\circ - \angle AOB$ (∠s at a pt).

Example 10: An unknown reflex angle

In the figure, $\angle AOB = 70^\circ$ , $\angle BOC = 95^\circ$ and reflex $\angle AOC = x^\circ$ . Find $x$ .

Solution

$\angle AOB$ , $\angle BOC$ and reflex $\angle AOC$ go all the way round $O$ , with no gap and no overlap.

$70 + 95 + x = 360$ (∠s at a pt)

$165 + x = 360$

$x = 360 - 165 = 195$

Check: $70^\circ + 95^\circ + 195^\circ = 360^\circ$ , and $195^\circ$ is between $180^\circ$ and $360^\circ$ , as a reflex angle must be.

$x = 195$ : reflex $\angle AOC = 195^\circ$ .

Using Both Rules

Some figures need one rule and then the other, or the same rule twice. Find what you can first, write each step with its reason, and use each answer in the next step.

Example 11: Two steps

In the figure, $POQ$ is a straight line through the point $O$ . Find $x$ and $y$ .

Solution

Step 1: above the line, $\angle POR$ and $\angle ROQ$ are adjacent angles on the straight line $POQ$ .

$4x + 5x = 180$ (adj. ∠s on a str. line)

$9x = 180$ , so $x = 20$ .

So $\angle POR = 4 \times 20^\circ = 80^\circ$ , $\angle ROQ = 5 \times 20^\circ = 100^\circ$ and $\angle QOS = 3 \times 20^\circ = 60^\circ$ .

Step 2: the four angles go all the way round $O$ , with no gap and no overlap.

$80 + 100 + 60 + y = 360$ (∠s at a pt)

$240 + y = 360$ , so $y = 120$ .

Check: below the line, $\angle QOS$ and $\angle SOP$ are adjacent angles on the straight line $POQ$ : $60^\circ + 120^\circ = 180^\circ$ (adj. ∠s on a str. line).

$x = 20$ and $y = 120$ : $\angle SOP = 120^\circ$ .

Common Mistakes

Using the wrong total. Angles on a straight line make a half turn, $180^\circ$ ; angles all the way round a point make a complete turn, $360^\circ$ . Before you write the equation, ask which one the angles make.
Adding angles that overlap. If $\angle AOC$ is made up of $\angle AOB$ and $\angle BOC$ , do not add all three. Add only angles that sit side by side, with no gap and no overlap.
Trusting your eyes. Use the straight-line rule only when the question tells you the line is straight, or you have shown it (Example 6). A line that looks straight in a figure may bend by a degree.
Leaving out the reflex angle. The angles at a point include the large angle on the outside, such as reflex $\angle AOC$ in Example 10.
Stopping at $x$ . If the angles are $2x^\circ$ and $3x^\circ$ and $x = 36$ , the angles are $72^\circ$ and $108^\circ$ , not $36^\circ$ . Read what the question asks for.
Dividing by the number of angles in a ratio question. For the ratio $4 : 5 : 6$ , divide $360$ by $4 + 5 + 6 = 15$ parts, not by $3$ .

Practice Questions

Give a reason for every step that uses an angle fact, and check that your angles add up to the right total. The figures are drawn to scale, but find each answer by working, not by measuring.

Question 1

In the figure, $AOB$ is a straight line. Find $p$ .

ShowHide solution

$p + 125 = 180$ (adj. ∠s on a str. line)

$p = 180 - 125 = 55$

Check: $55^\circ + 125^\circ = 180^\circ$ .

$p = 55$ , so $\angle COB = 55^\circ$ .

Question 2

In the figure, five arms meet at the point $O$ . Find $a$ .

ShowHide solution

The square marks a right angle: $\angle DOE = 90^\circ$ .

$60 + 75 + 85 + 90 + a = 360$ (∠s at a pt)

$310 + a = 360$

$a = 50$

Check: $60^\circ + 75^\circ + 85^\circ + 90^\circ + 50^\circ = 360^\circ$ .

$a = 50$ , so $\angle EOA = 50^\circ$ .

Question 3

In the figure, five arms meet at the point $O$ . The three angles marked $b^\circ$ are equal. Find $b$ .

ShowHide solution

$96 + 120 + b + b + b = 360$ (∠s at a pt)

$3b + 216 = 360$

Subtract $216$ from both sides: $3b = 144$

Divide both sides by $3$ : $b = 48$

Check: $96^\circ + 120^\circ + 48^\circ + 48^\circ + 48^\circ = 360^\circ$ .

$b = 48$ : each of the three equal angles is $48^\circ$ .

Question 4

In the figure, $\angle AOB = 70^\circ$ and $\angle BOC = 50^\circ$ .

Ravi writes: " $70^\circ + 50^\circ + \angle AOC = 360^\circ$ (∠s at a pt), so $\angle AOC = 240^\circ$ ."

(a) Explain what is wrong with Ravi's working.

(b) Find $\angle AOC$ and reflex $\angle AOC$ .

ShowHide solution

(a) $\angle AOC$ is made up of $\angle AOB$ and $\angle BOC$ , so Ravi's three angles overlap. The angles that go all the way round $O$ , with no gap and no overlap, are $\angle AOB$ , $\angle BOC$ and reflex $\angle AOC$ .

(b) $\angle AOC = \angle AOB + \angle BOC = 70^\circ + 50^\circ = 120^\circ$ .

Reflex $\angle AOC = 360^\circ - 70^\circ - 50^\circ = 240^\circ$ (∠s at a pt).

Check: $120^\circ + 240^\circ = 360^\circ$ . Ravi's $240^\circ$ is reflex $\angle AOC$ , not $\angle AOC$ .

(a) His three angles overlap, so the rule does not apply to them. (b) $\angle AOC = 120^\circ$ and reflex $\angle AOC = 240^\circ$ .

Question 5: The ribs of an umbrella

An open umbrella, seen from above, has $8$ ribs equally spaced round its centre $O$ . The figure numbers the ribs $1$ to $8$ .

(a) Find the angle between two neighbouring ribs.

(b) Find $p$ and $q$ .

ShowHide solution

(a) The $8$ equal angles between neighbouring ribs go all the way round $O$ . Let each be $t^\circ$ .

$8t = 360$ (∠s at a pt)

$t = 45$ : each angle between neighbouring ribs is $45^\circ$ .

(b) Going from rib $1$ past ribs $2$ and $3$ to rib $4$ there are $3$ of these angles: $p = 3 \times 45 = 135$ .

$p + q = 360$ (∠s at a pt), so $q = 360 - 135 = 225$ .

Check: from rib $4$ on past ribs $5$ , $6$ , $7$ and $8$ back to rib $1$ there are $5$ spaces, and $5 \times 45^\circ = 225^\circ$ ; $135^\circ + 225^\circ = 360^\circ$ .

(a) $45^\circ$ (b) $p = 135$ and $q = 225$ : the angle between rib $1$ and rib $4$ is $135^\circ$ , and the reflex angle is $225^\circ$ .

Question 6

In the figure, $AOB$ is a straight line. Find $x$ , and the size of each angle.

ShowHide solution

$(3x + 20) + (2x + 10) = 180$ (adj. ∠s on a str. line)

Remove the brackets: $3x + 20 + 2x + 10 = 180$

Collect like terms: $5x + 30 = 180$

$5x = 150$ , so $x = 30$ .

$\angle AOC = (3 \times 30 + 20)^\circ = 110^\circ$ and $\angle COB = (2 \times 30 + 10)^\circ = 70^\circ$ .

Check: $110^\circ + 70^\circ = 180^\circ$ .

$x = 30$ ; $\angle AOC = 110^\circ$ and $\angle COB = 70^\circ$ .

Question 7

In the figure, three arms meet at the point $O$ . Find $x$ , and the size of each angle.

ShowHide solution

$x + 2x + (x + 60) = 360$ (∠s at a pt)

Collect like terms: $4x + 60 = 360$

$4x = 300$ , so $x = 75$ .

$\angle AOB = 75^\circ$ , $\angle BOC = 2 \times 75^\circ = 150^\circ$ and $\angle COA = (75 + 60)^\circ = 135^\circ$ .

Check: $75^\circ + 150^\circ + 135^\circ = 360^\circ$ .

$x = 75$ ; the angles are $75^\circ$ , $150^\circ$ and $135^\circ$ .

Question 8

In the figure, three arms meet at the point $O$ , and $\angle AOB : \angle BOC :$ reflex $\angle AOC = 1 : 2 : 6$ . Find the three angles.

ShowHide solution

Let $\angle AOB = x^\circ$ , $\angle BOC = 2x^\circ$ and reflex $\angle AOC = 6x^\circ$ .

$x + 2x + 6x = 360$ (∠s at a pt)

$9x = 360$ , so $x = 40$ .

$\angle AOB = 40^\circ$ , $\angle BOC = 80^\circ$ and reflex $\angle AOC = 6 \times 40^\circ = 240^\circ$ .

Check: $40^\circ + 80^\circ + 240^\circ = 360^\circ$ ; $40 : 80 : 240 = 1 : 2 : 6$ ; and $240^\circ$ is between $180^\circ$ and $360^\circ$ , as a reflex angle must be.

$\angle AOB = 40^\circ$ , $\angle BOC = 80^\circ$ and reflex $\angle AOC = 240^\circ$ .

Question 9: A footpath off a road

At a point $O$ on a straight road $AB$ , a straight footpath $OC$ leads off the road. The angle between the footpath and the road on one side, $\angle AOC$ , is three times as large as the angle on the other side, $\angle COB$ . Find both angles.

ShowHide solution

Let $\angle COB = x^\circ$ . Then $\angle AOC = 3x^\circ$ .

$3x + x = 180$ (adj. ∠s on a str. line)

$4x = 180$ , so $x = 45$ .

$\angle COB = 45^\circ$ and $\angle AOC = 3 \times 45^\circ = 135^\circ$ .

Check: $135^\circ + 45^\circ = 180^\circ$ , and $135 = 3 \times 45$ . The two angles are supplementary.

$\angle AOC = 135^\circ$ and $\angle COB = 45^\circ$ .

Question 10

In each figure, decide whether the points lie on a straight line, and give a reason.

(a) Are $A$ , $O$ and $B$ on a straight line?

(b) Are $P$ , $O$ and $Q$ on a straight line?

ShowHide solution

(a) The three adjacent angles between $OA$ and $OB$ : $57^\circ + 64^\circ + 58^\circ = 179^\circ$ .

They add up to $179^\circ$ , not $180^\circ$ , so $AOB$ is not a straight line: it bends by $1^\circ$ at $O$ .

(b) The square marks a right angle: $\angle ROS = 90^\circ$ . The three adjacent angles between $OP$ and $OQ$ : $35^\circ + 90^\circ + 55^\circ = 180^\circ$ .

They add up to exactly $180^\circ$ , so $POQ$ is a straight line.

(a) No: the angles add up to $179^\circ$ , not $180^\circ$ . (b) Yes: the angles add up to $180^\circ$ .

Question 11: A road junction

Four straight roads meet at a junction $O$ . The angles between neighbouring roads are shown in the figure. Find $x$ , and the size of each angle.

ShowHide solution

The four angles go all the way round the junction.

$(2x + 10) + (4x - 15) + 95 + (x + 25) = 360$ (∠s at a pt)

Collect like terms ( $2x + 4x + x = 7x$ and $10 - 15 + 95 + 25 = 115$ ): $7x + 115 = 360$

Subtract $115$ from both sides: $7x = 245$

Divide both sides by $7$ : $x = 35$

$\angle AOB = (2 \times 35 + 10)^\circ = 80^\circ$ , $\angle BOC = (4 \times 35 - 15)^\circ = 125^\circ$ , $\angle COD = 95^\circ$ and $\angle DOA = (35 + 25)^\circ = 60^\circ$ .

Check: $80^\circ + 125^\circ + 95^\circ + 60^\circ = 360^\circ$ .

$x = 35$ ; the angles are $80^\circ$ , $125^\circ$ , $95^\circ$ and $60^\circ$ .

Question 12

In the figure, $AOB$ is a straight line and $\angle BOD$ is a right angle.

(a) Find $x$ , and the size of $\angle AOC$ and $\angle COB$ .

(b) Find reflex $\angle COD$ .

ShowHide solution

(a) $(4x - 12) + (x + 32) = 180$ (adj. ∠s on a str. line)

Collect like terms: $5x + 20 = 180$

$5x = 160$ , so $x = 32$ .

$\angle AOC = (4 \times 32 - 12)^\circ = 116^\circ$ and $\angle COB = (32 + 32)^\circ = 64^\circ$ . Check: $116^\circ + 64^\circ = 180^\circ$ .

(b) The angles that go all the way round $O$ are $\angle COB$ , $\angle BOD$ and reflex $\angle COD$ .

Reflex $\angle COD = 360^\circ - 64^\circ - 90^\circ = 206^\circ$ (∠s at a pt)

Check another way: $\angle AOD = 180^\circ - 90^\circ = 90^\circ$ (adj. ∠s on a str. line), and reflex $\angle COD = \angle COA + \angle AOD = 116^\circ + 90^\circ = 206^\circ$ .

(a) $x = 32$ ; $\angle AOC = 116^\circ$ and $\angle COB = 64^\circ$ . (b) Reflex $\angle COD = 206^\circ$ .

Key Points

Adjacent angles on a straight line add up to $180^\circ$ : reason (adj. ∠s on a str. line).
Angles at a point, which together make one complete turn with no gap and no overlap, add up to $360^\circ$ : reason (∠s at a pt).
Angles that meet at a point but do not go all the way round, or that overlap, are not covered by the rule.
Angles marked with the same letter are equal.
With angles in terms of $x$ : form the equation from the rule, solve it, find each angle, and check the total.
For angles in a ratio such as $4 : 5 : 6$ , write them as $4x^\circ$ , $5x^\circ$ and $6x^\circ$ .
Reflex $\angle AOB = 360^\circ - \angle AOB$ (∠s at a pt).
$AOB$ is a straight line exactly when the adjacent angles at $O$ between $OA$ and $OB$ add up to $180^\circ$ ; if they add up to any other number, it is not.
Write the reason after every step that uses an angle fact.

Before you start

Angles on a Straight Line

Angles on a straight line

Writing your working

Example 1: One unknown angle

Example 2: Three angles on a straight line

Example 3: Equal unknown angles

Adjacent, but not on a straight line

Forming and Solving an Equation

Example 4: Two angles in terms of xxx

Example 5: Expressions with brackets

Is It a Straight Line?

Testing for a straight line

Example 6: Do the points lie on a straight line?

Angles at a Point

Angles at a point

When the rule does not apply

Example 7: One unknown angle at a point

Example 8: Angles at a point in terms of xxx

Example 9: Angles in a given ratio

Example 10: An unknown reflex angle

Using Both Rules

Example 11: Two steps

Common Mistakes

Practice Questions

Question 1

Question 2

Question 3

Question 4

Question 5: The ribs of an umbrella

Question 6

Question 7

Question 8

Question 9: A footpath off a road

Question 10

Question 11: A road junction

Question 12

Key Points

Example 4: Two angles in terms of $x$

Example 8: Angles at a point in terms of $x$