AOB is a straight line and C is a point above the line. Angle AOC = 130°. Work out the size of angle COB.

Use the angle fact for a straight line: AOC + COB = 180° Subtract the known angle: COB = 180° - 130° = 50° State the answer: COB = 50° Answer: 50°

In triangle ABC, angle BAC = 40° and angle ABC = 75°. Sam writes ACB = 180 - 40 - 75 = 65. Which fact justifies Sam's working?

Look at the total Sam is using: 40° + 75° + 65° = 180° Name the fact: A + B + C = 180° Rule out the other reasons: no straight line, no crossing lines, no quadrilateral Answer: The angles in a triangle add up to 180°

In quadrilateral ABCD, angle DAB = 105°, angle ABC = 80° and angle BCD = 95°. Kim writes x = 360 - 105 - 80 - 95 = 80 for angle CDA = x°. Which fact justifies Kim's working?

Look at the total Kim starts from: 105 + 80 + 95 + 80 = 360 Name the fact: A + B + C + D = 360° Say where the 360 comes from: 2 × 180° = 360° Check the arithmetic: 360 - 105 - 80 - 95 = 80 Rule out the other reasons: no triangle, no crossing lines State the answer: CDA = 80° Answer: The angles in a quadrilateral add up to 360°

In triangle ABC, angle BAC = 55° and angle ABC = 48°. The side BC is extended to the point D. Ali writes ACD = 55 + 48 = 103. Which fact justifies Ali's working?

Say where the angle sits: ACD is outside the triangle Name the two interior opposite angles: BAC = 55°, ABC = 48° Name the fact: ACD = BAC + ABC Check the answer another way: ACB = 180° - 55° - 48° = 77° Rule out the interior angle next to it: ACB = 77° 103° Check the triangle is not isosceles: 55°, 48°, 77° Rule out the quadrilateral fact: there is no quadrilateral Rule out the false triangle statement: a triangle has 180°, not 360° Say which fact does the work: exterior angle = sum of the interior opposite angles State the answer: ACD = 103° Answer: The exterior angle equals the sum of the two interior opposite angles

ABC is a straight line with B between A and C, and D is a point above the line joined to B and to C. Angle DBA = (4x + 10)° and angle DBC = (2x + 20)°. In triangle BDC, angle BDC = (3x - 10)°. Work out the size of angle BCD.

Write down what you are given: DBA = (4x + 10)°, DBC = (2x + 20)°, BDC = (3x - 10)° State the fact that gives the equation: Angles on a straight line add up to 180° Form an equation: (4x + 10) + (2x + 20) = 180 Collect the like terms: 6x + 30 = 180 Subtract 30 from both sides: 6x = 150 Solve for x: x = 25 Find angle DBA: DBA = 4(25) + 10 = 110° Find angle DBC: DBC = 2(25) + 20 = 70° Check the straight line: 110 + 70 = 180 Find angle BDC: BDC = 3(25) - 10 = 65° State the fact for the next step: The angles in a triangle add up to 180° Work out angle BCD: BCD = 180° - 70° - 65° = 45° Check with the exterior angle fact: 65° + 45° = 110° Watch for the common error: BCD 180° - 110° - 65° State the answer: BCD = 45° Answer: 45°

GCSE Angle facts Practice Questions

Free GCSE Angle facts practice questions with full step-by-step worked solutions. Covers angles on a straight line, reasoning, three angles, angles at a point. Practise exam-style problems and check your method.

angles on a straight linereasoningthree anglesangles at a pointforming an equationvertically opposite angles

GCSE Foundation70 questionsStep-by-step solutions

Question 1

1 markeasy

AOB

is a straight line and

C

is a point above the line. Angle

AOC = 130^\circ

. Work out the size of angle

COB

Show worked solution

Worked solution

Use the angle fact for a straight line
$\angle AOC + \angle COB = 180^\circ$
Angles on a straight line add up to $180^{\circ}$ .
Subtract the known angle
$\angle COB = 180^\circ - 130^\circ = 50^\circ$
Taking $130^{\circ}$ away from $180^{\circ}$ leaves the angle you want.
State the answer
$\angle COB = 50^\circ$
Angle $COB$ measures $50^{\circ}$ .

Answer

50^\circ

Question 2

2 markseasy

In triangle

ABC

, angle

BAC = 40^\circ

and angle

ABC = 75^\circ

. Sam writes

\angle ACB = 180 - 40 - 75 = 65

. Which fact justifies Sam's working?

Show worked solution

Worked solution

Look at the total Sam is using
$40^\circ + 75^\circ + 65^\circ = 180^\circ$
Sam is sharing $180^{\circ}$ between the three angles of the triangle.
Name the fact
$\angle A + \angle B + \angle C = 180^\circ$
The angles in a triangle add up to $180^{\circ}$ .
Rule out the other reasons
$\text{no straight line, no crossing lines, no quadrilateral}$
The figure is a single triangle: nothing is extended into a straight line, no lines cross and there is no quadrilateral, so those facts have nothing to act on. The statement about $360^{\circ}$ in a triangle is simply false.

Answer

\text{The angles in a triangle add up to } 180^\circ

Question 3

2 marksintermediate

In quadrilateral

ABCD

, angle

DAB = 105^\circ

, angle

ABC = 80^\circ

and angle

BCD = 95^\circ

. Kim writes

x = 360 - 105 - 80 - 95 = 80

for angle

CDA = x^\circ

. Which fact justifies Kim's working?

Show worked solution

Worked solution

Look at the total Kim starts from
$105 + 80 + 95 + 80 = 360$
Kim is sharing $360^{\circ}$ between the four angles of the quadrilateral.
Name the fact
$\angle A + \angle B + \angle C + \angle D = 360^\circ$
The angles in a quadrilateral add up to $360^{\circ}$ .
Say where the $360$ comes from
$2 \times 180^\circ = 360^\circ$
A diagonal splits any quadrilateral into two triangles, and each triangle holds $180^{\circ}$ , so the four angles must total $360^{\circ}$ .
Check the arithmetic
$360 - 105 - 80 - 95 = 80$
The three known angles total $280^{\circ}$ , and $360$ minus $280$ is $80$ , so Kim has the right value.
Rule out the other reasons
$\text{no triangle, no crossing lines}$
The figure is a single quadrilateral: no lines cross, so there are no vertically opposite angles, and no triangle is drawn in it. The statements that a quadrilateral has $180^{\circ}$ and that a triangle has $360^{\circ}$ are both false.
State the answer
$\angle CDA = 80^\circ$
Kim is right, and the fact that justifies her working is that the angles in a quadrilateral add up to $360^{\circ}$ .

Answer

\text{The angles in a quadrilateral add up to } 360^\circ

Question 4

3 markshard

In triangle

ABC

, angle

BAC = 55^\circ

and angle

ABC = 48^\circ

. The side

BC

is extended to the point

D

. Ali writes

\angle ACD = 55 + 48 = 103

. Which fact justifies Ali's working?

Show worked solution

Worked solution

Say where the angle sits
$\angle ACD \text{ is outside the triangle}$
Angle $ACD$ lies outside triangle $ABC$ , between the side $CA$ and the extension $CD$ of the side $BC$ .
Name the two interior opposite angles
$\angle BAC = 55^\circ,\ \angle ABC = 48^\circ$
The interior opposite angles are the two angles of the triangle that are not next to the exterior angle.
Name the fact
$\angle ACD = \angle BAC + \angle ABC$
The exterior angle of a triangle is equal to the sum of the two interior opposite angles.
Check the answer another way
$\angle ACB = 180^\circ - 55^\circ - 48^\circ = 77^\circ$
The interior angle at $C$ is $77^\circ$ , and $B$ , $C$ and $D$ are on a straight line, so angle $ACD = 180 - 77 = 103^\circ$ . The two methods agree.
Rule out the interior angle next to it
$\angle ACB = 77^\circ \ne 103^\circ$
The interior angle next to the exterior angle is $77^{\circ}$ , not $103^{\circ}$ , so the claim that the exterior angle equals the interior angle beside it is false here.
Check the triangle is not isosceles
$55^\circ,\ 48^\circ,\ 77^\circ$
All three interior angles are different, so the triangle has no equal sides and the isosceles fact cannot justify anything in this figure.
Rule out the quadrilateral fact
$\text{there is no quadrilateral}$
The figure is a triangle with one side extended. There is no quadrilateral for a $360^{\circ}$ fact to act on.
Rule out the false triangle statement
$\text{a triangle has } 180^\circ,\ \text{not } 360^\circ$
A triangle has an angle sum of $180^{\circ}$ , so any statement giving it $360^{\circ}$ is false.
Say which fact does the work
$\text{exterior angle} = \text{sum of the interior opposite angles}$
Only the exterior angle fact turns $55 + 48$ straight into angle $ACD$ .
State the answer
$\angle ACD = 103^\circ$
The exterior angle $ACD$ measures $103^{\circ}$ .

Answer

\text{The exterior angle equals the sum of the two interior opposite angles}

Question 5

6 markschallenging

ABC

is a straight line with

B

between

A

and

C

, and

D

is a point above the line joined to

B

and to

C

. Angle

DBA = (4x + 10)^\circ

and angle

DBC = (2x + 20)^\circ

. In triangle

BDC

, angle

BDC = (3x - 10)^\circ

. Work out the size of angle

BCD

Show worked solution

Worked solution

Write down what you are given
$\angle DBA = (4x + 10)^\circ,\ \angle DBC = (2x + 20)^\circ,\ \angle BDC = (3x - 10)^\circ$
Angles $DBA$ and $DBC$ sit side by side on the straight line $ABC$ , so they are the pair that gives the equation.
State the fact that gives the equation
$\text{Angles on a straight line add up to } 180^\circ$
Angles on a straight line add up to $180^{\circ}$ .
Form an equation
$(4x + 10) + (2x + 20) = 180$
The two angles at $B$ fill the straight line $ABC$ , so together they make $180^{\circ}$ .
Collect the like terms
$6x + 30 = 180$
The $x$ terms give $4x + 2x = 6x$ , and the numbers give $10 + 20 = 30$ .
Subtract $30$ from both sides
$6x = 150$
Taking the number term off both sides leaves $6x$ on its own.
Solve for $x$
$x = 25$
Dividing both sides by $6$ gives $x = 25$ .
Find angle $DBA$
$\angle DBA = 4(25) + 10 = 110^\circ$
Substituting $x = 25$ into $4x + 10$ gives $110^\circ$ .
Find angle $DBC$
$\angle DBC = 2(25) + 20 = 70^\circ$
Substituting $x = 25$ into $2x + 20$ gives $70^\circ$ .
Check the straight line
$110 + 70 = 180$
The two angles at $B$ do add to $180^\circ$ , which confirms that $x = 25$ is right.
Find angle $BDC$
$\angle BDC = 3(25) - 10 = 65^\circ$
Substituting $x = 25$ into $3x - 10$ gives $65^\circ$ .
State the fact for the next step
$\text{The angles in a triangle add up to } 180^\circ$
The angles in a triangle add up to $180^{\circ}$ .
Work out angle $BCD$
$\angle BCD = 180^\circ - 70^\circ - 65^\circ = 45^\circ$
In triangle $BDC$ the angle at $B$ is $70^{\circ}$ and the angle at $D$ is $65^{\circ}$ , so subtract both from $180^{\circ}$ .
Check with the exterior angle fact
$65^\circ + 45^\circ = 110^\circ$
Angle $DBA = 110^{\circ}$ is the exterior angle of triangle $BDC$ at $B$ , so it must equal angle $BDC$ plus angle $BCD$ . It does, which checks every angle in the chain at once.
Watch for the common error
$\angle BCD \ne 180^\circ - 110^\circ - 65^\circ$
The $110^{\circ}$ angle lies outside triangle $BDC$ . The angle of the triangle at $B$ is $70^{\circ}$ , and that is the one that goes into the angle sum.
State the answer
$\angle BCD = 45^\circ$
Angle $BCD$ measures $45^{\circ}$ .

Answer

45^\circ

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