Hard GCSE Angle facts Questions

Challenging, exam-style GCSE Angle facts questions with worked solutions. Stretch yourself on the hardest angles on a straight line, forming an equation, solving equations, angles at a point problems.

angles on a straight lineforming an equationsolving equationsangles at a pointangles in a triangleexterior angle of a triangle
GCSE Foundation34 questionsStep-by-step solutions
Question 1
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
Question 2
6 markschallenging
In quadrilateral ABCDABCD, angle DAB=70∘DAB = 70^\circ, angle ABC=130∘ABC = 130^\circ and angle BCD=85∘BCD = 85^\circ. The diagonal BDBD is drawn, splitting the quadrilateral into triangle ABDABD and triangle BCDBCD. In triangle ABDABD, angle ADB=40∘ADB = 40^\circ. Work out the size of angle BDCBDC.
Show worked solution

Worked solution

  1. Write down what you are given

    ∠DAB=70∘, ∠ABC=130∘, ∠BCD=85∘, ∠ADB=40∘\angle DAB = 70^\circ,\ \angle ABC = 130^\circ,\ \angle BCD = 85^\circ,\ \angle ADB = 40^\circ

    List the angles you know before you start — each new angle must come from one of them.

  2. State the fact that gives angle ADC

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

    The angles in a quadrilateral add up to 360°.

  3. Work out angle ADC

    ∠ADC=360∘−70∘−130∘−85∘=75∘\angle ADC = 360^\circ - 70^\circ - 130^\circ - 85^\circ = 75^\circ

    The four angles of quadrilateral ABCD add to 360°, so the fourth one is what is left.

  4. State the fact that gives angle ABD

    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°.

  5. Work out angle ABD

    ∠ABD=180∘−70∘−40∘=70∘\angle ABD = 180^\circ - 70^\circ - 40^\circ = 70^\circ

    In triangle ABD the angles at A and D are known, so the third angle is 180° minus their total.

  6. State the fact that gives angle DBC

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

    The angles in a quadrilateral add up to 360°.

  7. Work out angle DBC

    ∠DBC=130∘−70∘=60∘\angle DBC = 130^\circ - 70^\circ = 60^\circ

    The diagonal BD splits the angle ABC of the quadrilateral into angle ABD and angle DBC, so subtract the part you have just found.

  8. State the fact that gives angle BDC

    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°.

  9. Work out angle BDC

    ∠BDC=180∘−85∘−60∘=35∘\angle BDC = 180^\circ - 85^\circ - 60^\circ = 35^\circ

    In triangle BCD the angles at C and B are now known, so subtract both from 180°.

  10. Check the two parts of angle ABC

    70+60=13070 + 60 = 130

    The diagonal BD splits angle ABC into angle ABD = 70° and angle DBC = 60°. Together they give back the 130° stated in the question.

  11. Check the two parts of angle ADC

    40+35=7540 + 35 = 75

    The diagonal also splits angle ADC into angle ADB = 40° and angle BDC = 35°. These add to 75°, which is exactly the angle ADC found from the quadrilateral — a complete check on the whole chain.

  12. Check triangle ABD adds up

    70+70+40=18070 + 70 + 40 = 180

    The three angles of triangle ABD total 180°.

  13. Check triangle BCD adds up

    85+60+35=18085 + 60 + 35 = 180

    The three angles of triangle BCD total 180°.

  14. Watch for the common error

    ∠DBC≠130∘\angle DBC \ne 130^\circ

    The 130° at B belongs to the quadrilateral, not to triangle BCD. Only the part of it below the diagonal, angle DBC = 60°, goes into that triangle.

  15. State the size of angle BDC

    ∠BDC=35∘\angle BDC = 35^\circ

    Angle bdc measures 35°.

Answer
35∘35^\circ
Question 3
5 markschallenging
Triangle ABCABC is isosceles with CA=CBCA = CB, and angle ACB=44∘ACB = 44^\circ. The side CBCB is extended beyond BB to the point DD. Work out the size of angle ABDABD, giving a reason for every step.
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Worked solution

  1. Write down what you are given

    CA=CB, ∠ACB=44∘CA = CB,\ \angle ACB = 44^\circ

    List the angles you know before you start — each new angle must come from one of them.

  2. State the fact that gives the pair of equal angles

    The base angles of an isosceles triangle are equal\text{The base angles of an isosceles triangle are equal}

    The base angles of an isosceles triangle are equal.

  3. Work out the pair of equal angles

    ∠CAB=∠CBA\angle CAB = \angle CBA

    CA = CB, so triangle ABC is isosceles. The equal angles are opposite those two equal sides: angle CBA (opposite CA) and angle CAB (opposite CB). Angle ACB is the apex angle.

  4. State the fact that gives what is left for the two base angles

    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°.

  5. Work out what is left for the two base angles

    180∘−44∘=136∘180^\circ - 44^\circ = 136^\circ

    The apex angle uses 44° of the 180°, leaving 136° for the two equal base angles together.

  6. State the fact that gives each base angle

    The base angles of an isosceles triangle are equal\text{The base angles of an isosceles triangle are equal}

    The base angles of an isosceles triangle are equal.

  7. Work out each base angle

    ∠CBA=136∘÷2=68∘\angle CBA = 136^\circ \div 2 = 68^\circ

    The two base angles are equal, so each one is half of 136°.

  8. State the fact that gives angle ABD

    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°.

  9. Work out angle ABD

    ∠ABD=180∘−68∘=112∘\angle ABD = 180^\circ - 68^\circ = 112^\circ

    C, B and D lie on a straight line, so angles CBA and ABD add to 180°.

  10. State the fact that gives a check on angle ABD

    The exterior angle of a triangle is equal to the sum of the two interior opposite angles\text{The exterior angle of a triangle is equal to the sum of the two interior opposite angles}

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

  11. Work out a check on angle ABD

    44∘+68∘=112∘44^\circ + 68^\circ = 112^\circ

    Angle ABD is the exterior angle of triangle ABC at B, so it should equal the two interior opposite angles (44° at C and 68° at A) added together — and it does.

  12. Check the triangle adds up

    44+68+68=18044 + 68 + 68 = 180

    The three interior angles total 180°, and each is between 0° and 180°, so the triangle can be drawn.

  13. Check the straight line

    68+112=18068 + 112 = 180

    Angle CBA and angle ABD lie on the straight line CBD and do add to 180°.

  14. Watch for the common error

    ∠ABD≠180∘−44∘\angle ABD \ne 180^\circ - 44^\circ

    The angle next to angle ABD on the straight line is the base angle CBA = 68°, not the apex angle 44°. Using the apex angle gives 136°, which is the total of the two base angles.

  15. State the size of angle ABD

    ∠ABD=112∘\angle ABD = 112^\circ

    Angle abd measures 112°.

Answer
112∘112^\circ
Question 4
5 markschallenging
In triangle ABCABC, angle BAC=38∘BAC = 38^\circ and angle ABC=57∘ABC = 57^\circ. The side ACAC is extended beyond CC to the point DD. Maya writes ∠BCD=38+57=95\angle BCD = 38 + 57 = 95. Which fact justifies Maya's working?
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Worked solution

  1. Write down what you are given

    ∠BAC=38∘, ∠ABC=57∘\angle BAC = 38^\circ,\ \angle ABC = 57^\circ

    Two of the three interior angles of triangle ABC are given.

  2. Say where angle BCD sits

    ∠BCD is outside the triangle\angle BCD \text{ is outside the triangle}

    Angle BCD is between the side CB and the extension CD of the side AC, so it lies outside triangle ABC.

  3. Name the two interior opposite angles

    ∠BAC and ∠ABC\angle BAC \text{ and } \angle ABC

    The interior opposite angles are the two angles of the triangle that are not next to angle BCD.

  4. Name the fact Maya is using

    ∠BCD=∠BAC+∠ABC\angle BCD = \angle BAC + \angle ABC

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

  5. Carry out Maya's addition

    38+57=9538 + 57 = 95

    The exterior angle BCD is 95°.

  6. Find the interior angle at C

    ∠ACB=180∘−38∘−57∘=85∘\angle ACB = 180^\circ - 38^\circ - 57^\circ = 85^\circ

    The three interior angles of the triangle add to 180°.

  7. Check Maya another way

    180∘−85∘=95∘180^\circ - 85^\circ = 95^\circ

    A, C and D lie on a straight line, so angle ACB and angle BCD add to 180°. This gives 95° again, so Maya is right.

  8. Rule out the interior angle next to it

    ∠ACB=85∘≠95∘\angle ACB = 85^\circ \ne 95^\circ

    The interior angle next to the exterior angle is 85°, not 95°, so a claim that the exterior angle equals the interior angle beside it is false here.

  9. Check the triangle is not isosceles

    38∘, 57∘, 85∘38^\circ,\ 57^\circ,\ 85^\circ

    All three interior angles are different, so triangle ABC has no equal sides and the isosceles base-angle fact cannot justify anything in this figure.

  10. Rule out the quadrilateral fact

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

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

  11. Rule out the vertically opposite fact

    no two lines cross\text{no two lines cross}

    Nothing in the figure crosses: AC is extended, but no second straight line passes through C, so there is no vertically opposite pair of angles.

  12. Rule out the false triangle statement

    38+57+85=18038 + 57 + 85 = 180

    The angles of this triangle total 180°, not 360°, so a statement giving a triangle 360° is false.

  13. 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 38 + 57 straight into angle BCD in one step.

  14. Write the reason the mark is for

    the exterior angle of a triangle\text{the exterior angle of a triangle}

    In the exam the mark is for quoting this reason, not just for writing down 95.

  15. State the answer

    ∠BCD=95∘\angle BCD = 95^\circ

    Maya is right, and her working is justified by the exterior angle fact.

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
Two straight lines ACAC and BDBD cross at the point OO. A third ray OEOE is drawn between OAOA and OBOB, so that angle AOE=3x∘AOE = 3x^\circ, angle EOB=2x∘EOB = 2x^\circ and angle BOC=(x+30)∘BOC = (x + 30)^\circ. Work out the size of angle CODCOD.
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Worked solution

  1. Write down what you are given

    ∠AOE=3x∘, ∠EOB=2x∘, ∠BOC=(x+30)∘\angle AOE = 3x^\circ,\ \angle EOB = 2x^\circ,\ \angle BOC = (x + 30)^\circ

    The three angles AOE, EOB and BOC sit side by side and together fill the straight line AOC.

  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

    3x+2x+(x+30)=1803x + 2x + (x + 30) = 180

    A, O and C lie on a straight line, so the three angles above it total 180°.

  4. Collect the like terms

    6x+30=1806x + 30 = 180

    The x terms give 3x+2x+x=6x3x + 2x + x = 6x, and the only number is 30.

  5. Subtract 30 from both sides

    6x=1506x = 150

    Taking 30 from 180 leaves 150.

  6. Solve for x

    x=25x = 25

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

  7. Find angle AOE

    ∠AOE=3×25=75∘\angle AOE = 3 \times 25 = 75^\circ

    Substituting x=25x = 25 into 3x3x gives 75∘75^\circ.

  8. Find angle EOB

    ∠EOB=2×25=50∘\angle EOB = 2 \times 25 = 50^\circ

    Substituting x=25x = 25 into 2x2x gives 50∘50^\circ.

  9. Find angle BOC

    ∠BOC=25+30=55∘\angle BOC = 25 + 30 = 55^\circ

    Substituting x=25x = 25 into x+30x + 30 gives 55∘55^\circ.

  10. Check the straight line

    75+50+55=18075 + 50 + 55 = 180

    The three angles on the straight line AOC do total 180∘180^\circ, so x=25x = 25 is right.

  11. Find angle AOB

    ∠AOB=75∘+50∘=125∘\angle AOB = 75^\circ + 50^\circ = 125^\circ

    Angle AOB is made up of angle AOE and angle EOB together.

  12. State the fact for the last step

    Vertically opposite angles are equal\text{Vertically opposite angles are equal}

    Vertically opposite angles are equal.

  13. Work out angle COD

    ∠COD=∠AOB=125∘\angle COD = \angle AOB = 125^\circ

    Angle COD is on the opposite side of O from angle AOB, made by the same two straight lines AC and BD.

  14. Check on the straight line

    125+55=180125 + 55 = 180

    Angle AOB = 125° and angle BOC = 55° lie on the straight line AOC, and they do add to 180°.

  15. State the answer

    ∠COD=125∘\angle COD = 125^\circ

    Angle COD measures 125°.

Answer
125∘125^\circ

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