Vertically Opposite Angles

When two straight lines cross, the angles opposite each other are equal. See why, find every angle at a crossing, form and solve equations in $x$ , and work with three lines through one point.

Before you start

An angle is named with three letters, the vertex in the middle: $\angle AOC$ is the angle at $O$ between $OA$ and $OC$ (Lesson 11.1).
Adjacent angles on a straight line add up to 180°, written (adj. ∠s on a str. line). If one of two angles on a straight line is 130°, the other is $180^{\circ} - 130^{\circ} = 50^{\circ}$ (Lesson 11.2).
Angles at a point add up to 360°, written (∠s at a pt), when together they make a complete turn with no gap and no overlap (Lesson 11.2).
An equation such as $4x + 20 = 180$ is solved by doing the same to both sides: subtract 20 from both sides to get $4x = 160$ , then divide both sides by 4 to get $x = 40$ (Lesson 9.1).

When Two Straight Lines Cross

Two straight lines that cross at a point make four angles there. Each angle has two adjacent angles (its neighbours), which share an arm with it, and one angle opposite it, which shares only the vertex.

Two angles opposite each other at a crossing are called vertically opposite angles. "Vertically" here comes from vertex: the two angles share a vertex and face each other across it. It has nothing to do with up and down: a pair of vertically opposite angles may lie left and right of the vertex just as well as above and below it.

In the figure, the straight lines $AB$ and $CD$ cross at $O$ . $\angle AOC$ and $\angle BOD$ are one pair of vertically opposite angles, and $\angle AOD$ and $\angle BOC$ are the other pair. The arms of one angle of a pair carry straight on through $O$ as the arms of the other: $OA$ carries on as $OB$ , and $OC$ carries on as $OD$ .

Vertically opposite angles

When two straight lines cross, the vertically opposite angles are equal.

In the figure: $\angle AOC = \angle BOD$ and $\angle AOD = \angle BOC$ .

The reason is written (vert. opp. ∠s), short for "vertically opposite angles are equal".

Why vertically opposite angles are equal

Call the four angles in the figure $a$ , $b$ , $c$ and $d$ .

$a + b = 180^{\circ}$ (adj. ∠s on a str. line), because $\angle AOC$ and $\angle COB$ together make the straight line $AB$ .

$c + b = 180^{\circ}$ (adj. ∠s on a str. line), because $\angle BOD$ and $\angle COB$ together make the straight line $CD$ .

So $a = 180^{\circ} - b$ and $c = 180^{\circ} - b$ . Both are 180° take away the same angle $b$ , so $a = c$ .

In the same way, $a + b = 180^{\circ}$ on line $AB$ and $a + d = 180^{\circ}$ on line $CD$ , so $b = d$ .

Nothing here depends on the size of the angles: the argument uses only the fact that $AB$ and $CD$ are straight. So the rule holds for every pair of straight lines that cross.

At a crossing, then, the two angles of each opposite pair are equal, and any two neighbours add up to 180° because they lie together on one of the straight lines. All four angles add up to 360° (∠s at a pt), which makes a useful check.

Example 1: Naming the pairs in a figure

The straight lines $PQ$ and $RS$ cross at $T$ .

(a) Name the angle vertically opposite $\angle PTR$ .

(b) Name the other pair of vertically opposite angles.

Solution

(a) The arms of $\angle PTR$ are $TP$ and $TR$ . Carry each straight on through $T$ : $TP$ carries on as $TQ$ , and $TR$ carries on as $TS$ . The angle between $TQ$ and $TS$ is $\angle QTS$ .

(b) The two angles left are $\angle RTQ$ and $\angle PTS$ . $TR$ carries on as $TS$ and $TQ$ as $TP$ , so they are vertically opposite.

(c) $\angle RTQ$ shares the arm $TR$ with $\angle PTR$ , and together they make the straight angle $\angle PTQ$ on line $PQ$ . $\angle PTS$ shares the arm $TP$ with it, and together they make the straight angle $\angle RTS$ on line $RS$ .

The angle vertically opposite $\angle PTR$ is $\angle QTS$ ; the other pair is $\angle RTQ$ and $\angle PTS$ ; $\angle RTQ$ and $\angle PTS$ each lie on a straight line with $\angle PTR$ .

Finding the Angles at a Crossing

One angle at a crossing fixes all four. The angle opposite it is equal to it (vert. opp. ∠s), and each of its neighbours makes 180° with it (adj. ∠s on a str. line). Write the reason after every step.

Example 2: All four angles from one

The straight lines $AB$ and $CD$ cross at $O$ , and $\angle AOC = 130^{\circ}$ . Find $\angle BOD$ , $\angle AOD$ and $\angle BOC$ .

Solution

$\angle BOD = \angle AOC = 130^{\circ}$ (vert. opp. ∠s)

$\angle AOC$ and $\angle AOD$ lie together on the straight line $CD$ , so

$\angle AOD = 180^{\circ} - 130^{\circ} = 50^{\circ}$ (adj. ∠s on a str. line)

$\angle BOC = \angle AOD = 50^{\circ}$ (vert. opp. ∠s)

Check: $130^{\circ} + 50^{\circ} + 130^{\circ} + 50^{\circ} = 360^{\circ}$ (∠s at a pt), and the neighbours $\angle BOC$ and $\angle BOD$ on line $CD$ add up to $50^{\circ} + 130^{\circ} = 180^{\circ}$ .

$\angle BOD = 130^{\circ}$ , $\angle AOD = 50^{\circ}$ and $\angle BOC = 50^{\circ}$ .

Example 3: Two vertically opposite angles with a known total

The straight lines $AB$ and $CD$ cross at $O$ , and $\angle AOC + \angle BOD = 104^{\circ}$ . Find $x$ .

Solution

$\angle AOC$ and $\angle BOD$ are vertically opposite, so they are equal (vert. opp. ∠s). Two equal angles that add up to 104° are each half of it:

$\angle AOC = 104^{\circ} \div 2 = 52^{\circ}$

The angle marked $x^{\circ}$ is $\angle BOC$ , which lies with $\angle AOC$ on the straight line $AB$ :

$x = 180 - 52 = 128$ (adj. ∠s on a str. line)

Check: $52^{\circ} + 128^{\circ} = 180^{\circ}$ , and the four angles at $O$ are 52°, 128°, 52° and 128°, which add up to 360°.

$x = 128$ , so $\angle BOC = 128^{\circ}$ .

Vertically Opposite Angles with Algebra

When the angles are given as expressions in $x$ , the rules give an equation. First look at where the two angles are:

Opposite each other: the two expressions are equal (vert. opp. ∠s).

Side by side on one straight line: the two expressions add up to 180 (adj. ∠s on a str. line).

Solve the equation by doing the same to both sides, put the value of $x$ back into every expression to find the angles, and check them.

Example 4: Vertically opposite angles in $x$

The straight lines $AB$ and $CD$ cross at $O$ . $\angle BOC = (3x + 10)^{\circ}$ and $\angle AOD = (5x - 30)^{\circ}$ . Find $x$ and the size of $\angle BOC$ .

Solution

$\angle BOC$ and $\angle AOD$ are vertically opposite, so

$5x - 30 = 3x + 10$ (vert. opp. ∠s)

Subtract $3x$ from both sides: $2x - 30 = 10$

Add 30 to both sides: $2x = 40$

Divide both sides by 2: $x = 20$

$\angle BOC = (3 \times 20 + 10)^{\circ} = 70^{\circ}$

Check: $\angle AOD = (5 \times 20 - 30)^{\circ} = 70^{\circ}$ . The two angles are equal, as they must be.

$x = 20$ and $\angle BOC = 70^{\circ}$ .

Example 5: Neighbouring angles in $x$

The straight lines $AB$ and $CD$ cross at $O$ . $\angle AOC = (2x + 15)^{\circ}$ and $\angle BOC = (3x - 35)^{\circ}$ . Find $x$ and all four angles at $O$ .

Solution

$\angle AOC$ and $\angle BOC$ lie together on the straight line $AB$ , so they add up to 180°:

$(2x + 15) + (3x - 35) = 180$ (adj. ∠s on a str. line)

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

Add 20 to both sides: $5x = 200$

Divide both sides by 5: $x = 40$

$\angle AOC = (2 \times 40 + 15)^{\circ} = 95^{\circ}$ and $\angle BOC = (3 \times 40 - 35)^{\circ} = 85^{\circ}$

$\angle BOD = \angle AOC = 95^{\circ}$ (vert. opp. ∠s)

$\angle AOD = \angle BOC = 85^{\circ}$ (vert. opp. ∠s)

Check: $95^{\circ} + 85^{\circ} = 180^{\circ}$ , and $95^{\circ} + 85^{\circ} + 95^{\circ} + 85^{\circ} = 360^{\circ}$ (∠s at a pt).

$x = 40$ ; $\angle AOC = \angle BOD = 95^{\circ}$ and $\angle BOC = \angle AOD = 85^{\circ}$ .

Three Lines Through One Point

When three straight lines pass through one point, they make six angles there. On each side of any one of the lines lie three of the angles, and they add up to 180° (adj. ∠s on a str. line). All six add up to 360° (∠s at a pt).

The six angles make three pairs of vertically opposite angles, one pair for each two of the lines. Two angles are vertically opposite only when both arms of one carry straight on as the arms of the other. In the figure of Example 6 below, $\angle BOC$ and $\angle AOD$ are vertically opposite ( $OB$ carries on as $OA$ , $OC$ as $OD$ ). But $\angle BOC$ and $\angle EOA$ are not, although they lie on either side of $O$ : $OC$ carries on as $OD$ , not as $OE$ .

Example 6: Six angles at a point

The straight lines $AB$ , $CD$ and $EF$ meet at $O$ . $\angle BOC = 70^{\circ}$ and $\angle COE = 60^{\circ}$ . Find $x$ , $y$ and $z$ , giving a reason for each step.

Solution

$\angle BOC$ , $\angle COE$ and the angle marked $x^{\circ}$ lie together on the straight line $AB$ , so

$x = 180 - 70 - 60 = 50$ (adj. ∠s on a str. line)

The angle marked $y^{\circ}$ is $\angle AOD$ . Its arms $OA$ and $OD$ carry straight on as $OB$ and $OC$ , so it is vertically opposite $\angle BOC$ :

$y = 70$ (vert. opp. ∠s)

The angle marked $z^{\circ}$ is $\angle DOF$ , vertically opposite $\angle COE$ :

$z = 60$ (vert. opp. ∠s)

Check: the last angle, $\angle FOB$ , is vertically opposite $\angle EOA$ , so it is 50°. All six: $70^{\circ} + 60^{\circ} + 50^{\circ} + 70^{\circ} + 60^{\circ} + 50^{\circ} = 360^{\circ}$ (∠s at a pt).

$x = 50$ , $y = 70$ and $z = 60$ : $\angle EOA = 50^{\circ}$ , $\angle AOD = 70^{\circ}$ and $\angle DOF = 60^{\circ}$ .

Example 7: Three lines, with an equation

The straight lines $AB$ , $CD$ and $EF$ meet at $O$ . $\angle AOF = 40^{\circ}$ , $\angle EOC = 2x^{\circ}$ and $\angle COA = (x + 20)^{\circ}$ . Find $x$ .

Solution

Above the straight line $AB$ lie $\angle BOE$ , $\angle EOC$ and $\angle COA$ . $\angle BOE$ is not marked, so find it first. Its arms $OB$ and $OE$ carry straight on as $OA$ and $OF$ :

$\angle BOE = \angle AOF = 40^{\circ}$ (vert. opp. ∠s)

$40 + 2x + (x + 20) = 180$ (adj. ∠s on a str. line)

Collect like terms: $3x + 60 = 180$

Subtract 60 from both sides: $3x = 120$

Divide both sides by 3: $x = 40$

Check: $\angle EOC = 2 \times 40^{\circ} = 80^{\circ}$ and $\angle COA = (40 + 20)^{\circ} = 60^{\circ}$ . On line $AB$ , $40^{\circ} + 80^{\circ} + 60^{\circ} = 180^{\circ}$ .

$x = 40$ , so $\angle EOC = 80^{\circ}$ and $\angle COA = 60^{\circ}$ .

When the Rule Does Not Apply

The rule needs two straight lines. If a "line" bends at the point, the angles opposite each other need not be equal, however much the figure looks like a crossing.

A figure cannot be trusted by eye: check with the angles. Three points $A$ , $O$ and $B$ lie on a straight line when the angles at $O$ on one side of $AOB$ add up to exactly 180° (Lesson 11.2).

Example 8: A line that is not straight

The four rays $OA$ , $OB$ , $OC$ and $OD$ start at $O$ , with the angles shown. Wei Ming says, " $\angle AOC$ and $\angle BOD$ are vertically opposite angles, so they are equal." Is he right?

Solution

First check $C$ , $O$ and $D$ : $\angle COB + \angle BOD = 75^{\circ} + 105^{\circ} = 180^{\circ}$ , so $COD$ is a straight line.

Then check $A$ , $O$ and $B$ : $\angle AOC + \angle COB = 115^{\circ} + 75^{\circ} = 190^{\circ}$ . This is not 180°, so $AOB$ is not a straight line: $OA$ does not carry straight on as $OB$ .

So $\angle AOC$ and $\angle BOD$ are not vertically opposite angles, and the rule says nothing about them. They are not equal: one is 115° and the other 105°.

The four angles still add up to $115^{\circ} + 75^{\circ} + 105^{\circ} + 65^{\circ} = 360^{\circ}$ (∠s at a pt): that rule needs no straight lines.

No. $AOB$ is not a straight line ( $115^{\circ} + 75^{\circ} = 190^{\circ}$ , not 180°), so the two angles are not vertically opposite, and they are not equal.

Common Mistakes

Taking two neighbouring angles as equal. Only the angles opposite each other are equal; neighbours add up to 180°. Where two lines cross at 70° and 110°, the 70° angle equals the angle opposite it, not the one beside it.
Reading "vertically" as "one above the other". Vertically opposite angles share a vertex and face each other across it; they may lie left and right of it.
With three lines through a point, calling two angles vertically opposite just because they lie on either side of the point. Both arms of one must carry straight on as the arms of the other.
Using the rule when a "line" bends at the point. Make sure each line is straight before writing (vert. opp. ∠s).
Stopping at $x$ . When the question asks for an angle, put $x$ back into the expression: in Example 4, $x = 20$ is not the angle; $(3x + 10)^{\circ} = 70^{\circ}$ is.
Writing the reason as "(opposite angles)". Write (vert. opp. ∠s): "opposite angles" means something else in a four-sided figure such as a parallelogram.

Practice

Try each question on paper first, giving a reason for every step, then open its solution.

Practice 1: Naming the pairs

The straight lines $KL$ and $MN$ cross at $O$ .

(a) Name the angle vertically opposite $\angle KOM$ .

(b) Name the angle vertically opposite $\angle MOL$ .

ShowHide solution

(a) $OK$ carries straight on as $OL$ , and $OM$ carries on as $ON$ . So the angle opposite $\angle KOM$ is $\angle LON$ .

(b) $OM$ carries on as $ON$ , and $OL$ carries on as $OK$ . So the angle opposite $\angle MOL$ is $\angle KON$ .

(c) $\angle MOL$ shares the arm $OM$ with $\angle KOM$ , and together they make the straight line $KL$ . $\angle KON$ shares the arm $OK$ with it, and together they make the straight line $MN$ .

(a) $\angle LON$ (b) $\angle KON$ (c) $\angle MOL$ and $\angle KON$

Practice 2: All four angles from one

The straight lines $AB$ and $CD$ cross at $O$ , and $\angle AOD = 72^{\circ}$ . Find $x$ , $y$ and $z$ .

ShowHide solution

The angle marked $x^{\circ}$ is $\angle BOC$ , vertically opposite $\angle AOD$ :

$x = 72$ (vert. opp. ∠s)

The angle marked $y^{\circ}$ is $\angle AOC$ , which lies with $\angle AOD$ on the straight line $CD$ :

$y = 180 - 72 = 108$ (adj. ∠s on a str. line)

The angle marked $z^{\circ}$ is $\angle BOD$ , vertically opposite $\angle AOC$ :

$z = 108$ (vert. opp. ∠s)

Check: $72^{\circ} + 72^{\circ} + 108^{\circ} + 108^{\circ} = 360^{\circ}$ (∠s at a pt).

$x = 72$ , $y = 108$ and $z = 108$ : $\angle BOC = 72^{\circ}$ and $\angle AOC = \angle BOD = 108^{\circ}$ .

Practice 3: A pair of scissors

A pair of scissors is made of two straight pieces of metal joined by a screw at $O$ . Each piece is a blade at one end and a handle at the other: one piece is $PS$ and the other is $QR$ . The handles $OR$ and $OS$ are opened to an angle of 38°.

(a) Find $\angle POQ$ , the angle between the blades.

(b) Find $\angle POR$ , the angle between the blade $OP$ and the handle $OR$ .

ShowHide solution

(a) $PS$ and $QR$ are straight lines crossing at $O$ , so $\angle POQ$ and $\angle ROS$ are vertically opposite:

$\angle POQ = \angle ROS = 38^{\circ}$ (vert. opp. ∠s)

(b) $\angle POR$ and $\angle ROS$ lie together on the straight line $PS$ :

$\angle POR = 180^{\circ} - 38^{\circ} = 142^{\circ}$ (adj. ∠s on a str. line)

Check: the four angles at $O$ are 38°, 142°, 38° and 142°, which add up to 360°.

(a) $\angle POQ = 38^{\circ}$ : the blades open exactly as wide as the handles. (b) $\angle POR = 142^{\circ}$ .

Practice 4: A known total

The straight lines $AB$ and $CD$ cross at $O$ , and $\angle BOC + \angle AOD = 136^{\circ}$ . Find $x$ .

ShowHide solution

$\angle BOC$ and $\angle AOD$ are vertically opposite, so they are equal (vert. opp. ∠s). Each is half of 136°:

$\angle BOC = 136^{\circ} \div 2 = 68^{\circ}$

The angle marked $x^{\circ}$ is $\angle AOC$ , which lies with $\angle BOC$ on the straight line $AB$ :

$x = 180 - 68 = 112$ (adj. ∠s on a str. line)

Check: $68^{\circ} + 112^{\circ} = 180^{\circ}$ .

$x = 112$ , so $\angle AOC = 112^{\circ}$ .

Practice 5: Equal expressions

Two straight lines cross. Two vertically opposite angles are $(4x - 15)^{\circ}$ and $(x + 36)^{\circ}$ , as shown. Find $x$ and the size of each of these angles.

ShowHide solution

$4x - 15 = x + 36$ (vert. opp. ∠s)

Subtract $x$ from both sides: $3x - 15 = 36$

Add 15 to both sides: $3x = 51$

Divide both sides by 3: $x = 17$

$(4x - 15)^{\circ} = (4 \times 17 - 15)^{\circ} = 53^{\circ}$

Check: $(x + 36)^{\circ} = (17 + 36)^{\circ} = 53^{\circ}$ . The two angles are equal.

$x = 17$ , and each of the two angles is $53^{\circ}$ .

Practice 6: Neighbouring expressions

The straight lines $AB$ and $CD$ cross at $O$ . $\angle AOC = 5x^{\circ}$ and $\angle BOC = (2x + 12)^{\circ}$ . Find $x$ and all four angles at $O$ .

ShowHide solution

$\angle AOC$ and $\angle BOC$ lie together on the straight line $AB$ :

$5x + (2x + 12) = 180$ (adj. ∠s on a str. line)

Collect like terms: $7x + 12 = 180$

Subtract 12 from both sides: $7x = 168$

Divide both sides by 7: $x = 24$

$\angle AOC = 5 \times 24^{\circ} = 120^{\circ}$ and $\angle BOC = (2 \times 24 + 12)^{\circ} = 60^{\circ}$

$\angle BOD = \angle AOC = 120^{\circ}$ (vert. opp. ∠s)

$\angle AOD = \angle BOC = 60^{\circ}$ (vert. opp. ∠s)

Check: $120^{\circ} + 60^{\circ} + 120^{\circ} + 60^{\circ} = 360^{\circ}$ (∠s at a pt).

$x = 24$ ; $\angle AOC = \angle BOD = 120^{\circ}$ and $\angle BOC = \angle AOD = 60^{\circ}$ .

Practice 7: Two steps

The straight lines $AB$ and $CD$ cross at $O$ . $\angle BOC = 3x^{\circ}$ , $\angle AOD = (x + 64)^{\circ}$ and $\angle AOC = y^{\circ}$ . Find $x$ and $y$ .

ShowHide solution

$\angle BOC$ and $\angle AOD$ are vertically opposite:

$3x = x + 64$ (vert. opp. ∠s)

Subtract $x$ from both sides: $2x = 64$

Divide both sides by 2: $x = 32$

So $\angle BOC = 3 \times 32^{\circ} = 96^{\circ}$ , and $\angle AOD = (32 + 64)^{\circ} = 96^{\circ}$ too.

$\angle AOC$ lies with $\angle BOC$ on the straight line $AB$ :

$y = 180 - 96 = 84$ (adj. ∠s on a str. line)

Check: $96^{\circ} + 84^{\circ} = 180^{\circ}$ .

$x = 32$ and $y = 84$ : $\angle BOC = \angle AOD = 96^{\circ}$ and $\angle AOC = 84^{\circ}$ .

Practice 8: A folding table

The two legs of a folding table, $AB$ and $CD$ , are straight and cross at $O$ . The angle between the legs under the tabletop, $\angle AOC$ , is $2x^{\circ}$ , and the angle on each side, such as $\angle COB$ , is $(x + 30)^{\circ}$ . Find $x$ , and $\angle BOD$ , the angle between the legs at the floor.

ShowHide solution

$\angle AOC$ and $\angle COB$ lie together on the straight leg $AB$ :

$2x + (x + 30) = 180$ (adj. ∠s on a str. line)

Collect like terms: $3x + 30 = 180$

Subtract 30 from both sides: $3x = 150$

Divide both sides by 3: $x = 50$

$\angle AOC = 2 \times 50^{\circ} = 100^{\circ}$

$\angle BOD = \angle AOC = 100^{\circ}$ (vert. opp. ∠s)

Check: $\angle COB = (50 + 30)^{\circ} = 80^{\circ}$ , and $100^{\circ} + 80^{\circ} = 180^{\circ}$ .

$x = 50$ and $\angle BOD = 100^{\circ}$ : the legs open as wide at the floor as under the tabletop.

Practice 9: Three lines

The straight lines $AB$ , $CD$ and $EF$ meet at $O$ . $\angle BOE = 48^{\circ}$ and $\angle FOD = 75^{\circ}$ . Find $x$ , $y$ and $z$ .

ShowHide solution

$\angle EOC$ is not marked. Its arms $OE$ and $OC$ carry straight on as $OF$ and $OD$ , so it is vertically opposite $\angle FOD$ :

$\angle EOC = 75^{\circ}$ (vert. opp. ∠s)

The angle marked $x^{\circ}$ is $\angle COA$ . On the straight line $AB$ :

$x = 180 - 48 - 75 = 57$ (adj. ∠s on a str. line)

The angle marked $y^{\circ}$ is $\angle AOF$ , vertically opposite $\angle BOE$ :

$y = 48$ (vert. opp. ∠s)

The angle marked $z^{\circ}$ is $\angle DOB$ , vertically opposite $\angle COA$ :

$z = 57$ (vert. opp. ∠s)

Check: below $AB$ , $48^{\circ} + 75^{\circ} + 57^{\circ} = 180^{\circ}$ (adj. ∠s on a str. line).

$x = 57$ , $y = 48$ and $z = 57$ : $\angle COA = \angle DOB = 57^{\circ}$ and $\angle AOF = 48^{\circ}$ .

Practice 10: Three lines, with an equation

The straight lines $AB$ , $CD$ and $EF$ meet at $O$ . $\angle BOE = 3x^{\circ}$ , $\angle EOC = 50^{\circ}$ and $\angle AOF = (x + 30)^{\circ}$ . Find $x$ and $\angle AOC$ .

ShowHide solution

$\angle BOE$ and $\angle AOF$ are vertically opposite ( $OB$ carries on as $OA$ , $OE$ as $OF$ ):

$3x = x + 30$ (vert. opp. ∠s)

Subtract $x$ from both sides: $2x = 30$

Divide both sides by 2: $x = 15$

So $\angle BOE = 3 \times 15^{\circ} = 45^{\circ}$ , and $\angle AOF = (15 + 30)^{\circ} = 45^{\circ}$ too.

$\angle BOE$ , $\angle EOC$ and $\angle AOC$ lie together on the straight line $AB$ :

$\angle AOC = 180^{\circ} - 45^{\circ} - 50^{\circ} = 85^{\circ}$ (adj. ∠s on a str. line)

Check: $45^{\circ} + 50^{\circ} + 85^{\circ} = 180^{\circ}$ .

$x = 15$ and $\angle AOC = 85^{\circ}$ .

Practice 11: Is it a crossing?

The four rays $OA$ , $OB$ , $OC$ and $OD$ start at $O$ . $\angle AOC = 66^{\circ}$ , $\angle COB = 120^{\circ}$ and $\angle BOD = 60^{\circ}$ .

(a) Find $\angle AOD$ .

(b) Which of $AOB$ and $COD$ is a straight line?

ShowHide solution

(a) The four angles make a complete turn about $O$ :

$\angle AOD = 360^{\circ} - 66^{\circ} - 120^{\circ} - 60^{\circ} = 114^{\circ}$ (∠s at a pt)

(b) $\angle COB + \angle BOD = 120^{\circ} + 60^{\circ} = 180^{\circ}$ , so $COD$ is a straight line.

$\angle AOC + \angle COB = 66^{\circ} + 120^{\circ} = 186^{\circ}$ . This is not 180°, so $AOB$ is not a straight line, although it looks nearly straight.

(c) Vertically opposite angles are equal only when both lines are straight. $AOB$ bends at $O$ , so $\angle AOC$ and $\angle BOD$ are not vertically opposite angles. They are not equal: one is 66° and the other 60°.

(a) $\angle AOD = 114^{\circ}$ (b) $COD$ is a straight line; $AOB$ is not. (c) $AOB$ is not straight, so the angles are not vertically opposite; they are 66° and 60°, not equal.

Practice 12: Paths in a park

Three straight paths in a park, $AB$ , $CD$ and $EF$ , cross at one point $O$ . Three of the angles between them are marked. Find $x$ and all six angles at $O$ .

ShowHide solution

The three marked angles lie together on the straight path $AB$ :

$x + 2x + (x + 40) = 180$ (adj. ∠s on a str. line)

Collect like terms: $4x + 40 = 180$

Subtract 40 from both sides: $4x = 140$

Divide both sides by 4: $x = 35$

So $\angle BOE = 35^{\circ}$ , $\angle EOC = 2 \times 35^{\circ} = 70^{\circ}$ and $\angle COA = (35 + 40)^{\circ} = 75^{\circ}$ .

Each angle below $AB$ is vertically opposite one of these (vert. opp. ∠s):

$\angle AOF = \angle BOE = 35^{\circ}$ , $\angle FOD = \angle EOC = 70^{\circ}$ and $\angle DOB = \angle COA = 75^{\circ}$ .

Check: $35^{\circ} + 70^{\circ} + 75^{\circ} + 35^{\circ} + 70^{\circ} + 75^{\circ} = 360^{\circ}$ (∠s at a pt).

$x = 35$ ; $\angle BOE = \angle AOF = 35^{\circ}$ , $\angle EOC = \angle FOD = 70^{\circ}$ and $\angle COA = \angle DOB = 75^{\circ}$ .

Key Points

Two straight lines that cross make four angles: two pairs of vertically opposite angles.
Vertically opposite angles are equal (vert. opp. ∠s).
Neighbouring angles at a crossing lie on a straight line and add up to 180° (adj. ∠s on a str. line).
The angles all the way round the crossing add up to 360° (∠s at a pt): use it as a check.
One angle at a crossing gives all four.
With expressions in $x$ : vertically opposite expressions are equal; neighbouring expressions add up to 180. Solve, find each angle, and check.
With three lines through a point, two angles are vertically opposite only when both arms of one carry straight on as the arms of the other.
The rule needs straight lines: if the angles along a "line" do not add up to 180°, it bends, and the rule does not apply.