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
AOBAOB is a straight line and CC is a point above the line. Angle AOC=130∘AOC = 130^\circ. Work out the size of angle COBCOB.
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Worked solution

  1. Use the angle fact for a straight line

    ∠AOC+∠COB=180∘\angle AOC + \angle COB = 180^\circ

    Angles on a straight line add up to 180°.

  2. Subtract the known angle

    ∠COB=180∘−130∘=50∘\angle COB = 180^\circ - 130^\circ = 50^\circ

    Taking 130° away from 180° leaves the angle you want.

  3. State the answer

    ∠COB=50∘\angle COB = 50^\circ

    Angle COB measures 50°.

Answer
50∘50^\circ
Question 2
2 markseasy
In triangle ABCABC, angle BAC=40∘BAC = 40^\circ and angle ABC=75∘ABC = 75^\circ. Sam writes ∠ACB=180−40−75=65\angle ACB = 180 - 40 - 75 = 65. Which fact justifies Sam's working?
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Worked solution

  1. Look at the total Sam is using

    40∘+75∘+65∘=180∘40^\circ + 75^\circ + 65^\circ = 180^\circ

    Sam is sharing 180° between the three angles of the triangle.

  2. Name the fact

    ∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^\circ

    The angles in a triangle add up to 180°.

  3. Rule out the other reasons

    no straight line, no crossing lines, no quadrilateral\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° in a triangle is simply false.

Answer
The angles in a triangle add up to 180∘\text{The angles in a triangle add up to } 180^\circ
Question 3
2 marksintermediate
In quadrilateral ABCDABCD, angle DAB=105∘DAB = 105^\circ, angle ABC=80∘ABC = 80^\circ and angle BCD=95∘BCD = 95^\circ. Kim writes x=360−105−80−95=80x = 360 - 105 - 80 - 95 = 80 for angle CDA=x∘CDA = x^\circ. Which fact justifies Kim's working?
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Worked solution

  1. Look at the total Kim starts from

    105+80+95+80=360105 + 80 + 95 + 80 = 360

    Kim is sharing 360° between the four angles of the quadrilateral.

  2. Name the fact

    ∠A+∠B+∠C+∠D=360∘\angle A + \angle B + \angle C + \angle D = 360^\circ

    The angles in a quadrilateral add up to 360°.

  3. Say where the 360 comes from

    2×180∘=360∘2 \times 180^\circ = 360^\circ

    A diagonal splits any quadrilateral into two triangles, and each triangle holds 180°, so the four angles must total 360°.

  4. Check the arithmetic

    360−105−80−95=80360 - 105 - 80 - 95 = 80

    The three known angles total 280°, and 360 minus 280 is 80, so Kim has the right value.

  5. Rule out the other reasons

    no triangle, no crossing lines\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° and that a triangle has 360° are both false.

  6. State the answer

    ∠CDA=80∘\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°.

Answer
The angles in a quadrilateral add up to 360∘\text{The angles in a quadrilateral add up to } 360^\circ
Question 4
3 markshard
In triangle ABCABC, angle BAC=55∘BAC = 55^\circ and angle ABC=48∘ABC = 48^\circ. The side BCBC is extended to the point DD. Ali writes ∠ACD=55+48=103\angle ACD = 55 + 48 = 103. Which fact justifies Ali's working?
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Worked solution

  1. Say where the angle sits

    ∠ACD is outside the triangle\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.

  2. Name the two interior opposite angles

    ∠BAC=55∘, ∠ABC=48∘\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.

  3. Name the fact

    ∠ACD=∠BAC+∠ABC\angle ACD = \angle BAC + \angle ABC

    The exterior angle of a triangle is equal to the sum of the two interior opposite angles.

  4. Check the answer another way

    ∠ACB=180∘−55∘−48∘=77∘\angle ACB = 180^\circ - 55^\circ - 48^\circ = 77^\circ

    The interior angle at C is 77∘77^\circ, and B, C and D are on a straight line, so angle ACD=180−77=103∘ACD = 180 - 77 = 103^\circ. The two methods agree.

  5. Rule out the interior angle next to it

    ∠ACB=77∘≠103∘\angle ACB = 77^\circ \ne 103^\circ

    The interior angle next to the exterior angle is 77°, not 103°, so the claim that the exterior angle equals the interior angle beside it is false here.

  6. Check the triangle is not isosceles

    55∘, 48∘, 77∘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.

  7. Rule out the quadrilateral fact

    there is no quadrilateral\text{there is no quadrilateral}

    The figure is a triangle with one side extended. There is no quadrilateral for a 360° fact to act on.

  8. Rule out the false triangle statement

    a triangle has 180∘, not 360∘\text{a triangle has } 180^\circ,\ \text{not } 360^\circ

    A triangle has an angle sum of 180°, so any statement giving it 360° is false.

  9. Say which fact does the work

    exterior angle=sum of the interior opposite angles\text{exterior angle} = \text{sum of the interior opposite angles}

    Only the exterior angle fact turns 55 + 48 straight into angle ACD.

  10. State the answer

    ∠ACD=103∘\angle ACD = 103^\circ

    The exterior angle ACD measures 103°.

Answer
The exterior angle equals the sum of the two interior opposite angles\text{The exterior angle equals the sum of the two interior opposite angles}
Question 5
6 markschallenging
ABCABC is a straight line with BB between AA and CC, and DD is a point above the line joined to BB and to CC. Angle DBA=(4x+10)∘DBA = (4x + 10)^\circ and angle DBC=(2x+20)∘DBC = (2x + 20)^\circ. In triangle BDCBDC, angle BDC=(3x−10)∘BDC = (3x - 10)^\circ. Work out the size of angle BCDBCD.
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Worked solution

  1. Write down what you are given

    ∠DBA=(4x+10)∘, ∠DBC=(2x+20)∘, ∠BDC=(3x−10)∘\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.

  2. State the fact that gives the equation

    Angles on a straight line add up to 180∘\text{Angles on a straight line add up to } 180^\circ

    Angles on a straight line add up to 180°.

  3. Form an equation

    (4x+10)+(2x+20)=180(4x + 10) + (2x + 20) = 180

    The two angles at B fill the straight line ABC, so together they make 180°.

  4. Collect the like terms

    6x+30=1806x + 30 = 180

    The x terms give 4x+2x=6x4x + 2x = 6x, and the numbers give 10+20=3010 + 20 = 30.

  5. Subtract 30 from both sides

    6x=1506x = 150

    Taking the number term off both sides leaves 6x on its own.

  6. Solve for x

    x=25x = 25

    Dividing both sides by 6 gives x=25x = 25.

  7. Find angle DBA

    ∠DBA=4(25)+10=110∘\angle DBA = 4(25) + 10 = 110^\circ

    Substituting x=25x = 25 into 4x+104x + 10 gives 110∘110^\circ.

  8. Find angle DBC

    ∠DBC=2(25)+20=70∘\angle DBC = 2(25) + 20 = 70^\circ

    Substituting x=25x = 25 into 2x+202x + 20 gives 70∘70^\circ.

  9. Check the straight line

    110+70=180110 + 70 = 180

    The two angles at B do add to 180∘180^\circ, which confirms that x=25x = 25 is right.

  10. Find angle BDC

    ∠BDC=3(25)−10=65∘\angle BDC = 3(25) - 10 = 65^\circ

    Substituting x=25x = 25 into 3x−103x - 10 gives 65∘65^\circ.

  11. State the fact for the next step

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

    The angles in a triangle add up to 180°.

  12. Work out angle BCD

    ∠BCD=180∘−70∘−65∘=45∘\angle BCD = 180^\circ - 70^\circ - 65^\circ = 45^\circ

    In triangle BDC the angle at B is 70° and the angle at D is 65°, so subtract both from 180°.

  13. Check with the exterior angle fact

    65∘+45∘=110∘65^\circ + 45^\circ = 110^\circ

    Angle DBA = 110° 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.

  14. Watch for the common error

    ∠BCD≠180∘−110∘−65∘\angle BCD \ne 180^\circ - 110^\circ - 65^\circ

    The 110° angle lies outside triangle BDC. The angle of the triangle at B is 70°, and that is the one that goes into the angle sum.

  15. State the answer

    ∠BCD=45∘\angle BCD = 45^\circ

    Angle BCD measures 45°.

Answer
45∘45^\circ

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