[discia] GCSE Maths › Geometry & Measures › Angles › Practise
  1. Either paper 3 marks

    Here are the sizes of four angles.

    \(332^\circ\) · \(90^\circ\) · \(106^\circ\) · \(32^\circ\)

    One of them is a right angle.

    (a)

    State the size of the angle that is acute.

    (b)

    State the size of the angle that is obtuse.

    (c)

    State the size of the angle that is reflex.

    Show answer
    (a) \(32^\circ\)
    (b) \(106^\circ\)
    (c) \(332^\circ\)

    An acute angle is smaller than a right angle — anything less than \(90^\circ\).

    Of the four sizes, only \(32^\circ\) is below \(90^\circ\).

    An obtuse angle is bigger than a right angle but smaller than a half turn: between \(90^\circ\) and \(180^\circ\).

    \(106^\circ\) is the only one in that range. \(90^\circ\) is a right angle, not an obtuse one, which is the answer this question is really testing.

    A reflex angle is bigger than a half turn: between \(180^\circ\) and \(360^\circ\).

    \(332^\circ\) is more than \(180^\circ\), so it is the reflex one.

  2. Either paper 2 marks

    \(AOB\) is a straight line.

    \(OC\) and \(OD\) are two more straight lines drawn from \(O\), both on the same side of \(AOB\), with \(OC\) between \(OA\) and \(OD\).

    Angle \(AOC = 91^\circ\) and angle \(COD = 69^\circ\).

    Work out the size of angle \(DOB\).

    Show answer
    \(20^\circ\)

    \(AOC\), \(COD\) and \(DOB\) are the three angles that fill the space above the straight line \(AOB\), so together they make a half turn.

    Angles on a straight line add up to \(180^\circ\).

    \(91 + 69 = 160\)

    \(180 - 160 = 20\)

    Angle \(DOB = 20^\circ\)

    • M1adds the two given angles to get \(160\) or uses 180 as the total
    • A1angle \(DOB = 20^\circ\)
  3. Either paper 2 marks

    \(OA\), \(OB\), \(OC\) and \(OD\) are four straight lines drawn from the point \(O\), arranged in that order all the way round \(O\).

    Angle \(AOB = 79^\circ\), angle \(BOC = 46^\circ\) and angle \(COD = 116^\circ\).

    Calculate the size of angle \(DOA\).

    Show answer
    \(119^\circ\)

    \(AOB\), \(BOC\), \(COD\) and \(DOA\) go once all the way round \(O\), so together they make a full turn.

    Angles at a point add up to \(360^\circ\).

    \(79 + 46 + 116 = 241\)

    \(360 - 241 = 119\)

    Angle \(DOA = 119^\circ\)

    • M1adds the three given angles to get \(241\) or uses 360 as the total
    • A1angle \(DOA = 119^\circ\)
  4. Either paper 2 marks

    The diagram shows a triangle.

    48°72°xDiagram NOT accurately drawn

    Calculate the size of the angle marked \(x\).

    Show answer
    \(60^\circ\)

    The three angles of any triangle add up to \(180^\circ\).

    Add the two you are given:

    \(48 + 72 = 120\)

    Take that away from \(180^\circ\):

    \(x = 180 - 120 = 60\)

    \(x = 60^\circ\)

    • M1adds the two given angles to get \(120\) or uses 180 as the total
    • A1\(x = 60^\circ\)
  5. Either paper 1 mark

    The diagram is drawn accurately.

    x

    Estimate the size of the angle marked \(x\).

    You must not use a protractor.

    Show answer
    \(50^\circ\) — accept any answer from \(40^\circ\) to \(60^\circ\)

    Compare the angle with the angles you already know the size of. A right angle is \(90^\circ\) and a straight line is \(180^\circ\), and halving a right angle gives \(45^\circ\).

    The angle marked \(x\) is bigger than half a right angle but still smaller than a right angle, so an estimate of about \(50^\circ\) is right.

    Anything from \(40^\circ\) to \(60^\circ\) would earn the mark. Estimating is not guessing: the marks are for landing close, and comparing with \(90^\circ\) and \(45^\circ\) first is what gets you there.

  6. Either paper 2 marks

    Here is a right-angled triangle.

    35°xDiagram NOT accurately drawn

    Find the size of the angle marked \(x\).

    Show answer
    \(55^\circ\)

    The square in the corner means that angle is \(90^\circ\).

    The three angles of a triangle add up to \(180^\circ\):

    \(90 + 35 = 125\)

    \(x = 180 - 125 = 55\)

    \(x = 55^\circ\)

    A useful shortcut worth remembering: in a right-angled triangle the two acute angles always add up to \(90^\circ\), so \(x = 90 - 35\) straight away.

    • M1uses \(90^\circ\) and \(35^\circ\) with a total of 180, e.g. \(90 + 35 = 125\)
    • A1\(x = 55^\circ\)
  7. Either paper 4 marks

    Two straight lines \(AB\) and \(CD\) cross at the point \(O\).

    Angle \(AOC = 140^\circ\).

    (a)

    Write down the size of angle \(BOD\), and give a reason for your answer.

    (b)

    Find the size of angle \(AOD\), and give a reason for your answer.

    Show answer
    (a) \(BOD = 140^\circ\), because vertically opposite angles are equal
    (b) \(AOD = 40^\circ\), because angles on a straight line add up to 180 degrees

    \(AOC\) and \(BOD\) are on opposite sides of \(O\): \(OB\) points the opposite way to \(OA\), and \(OD\) points the opposite way to \(OC\). That is what makes them a vertically opposite pair.

    Vertically opposite angles are equal, so angle \(BOD = 140^\circ\).

    "Vertically opposite" has nothing to do with vertical lines — it means opposite through the vertex.

    \(C\), \(O\) and \(D\) lie on one straight line, and \(OA\) is drawn from \(O\), so angle \(AOC\) and angle \(AOD\) together make a half turn.

    Angles on a straight line add up to \(180^\circ\), so

    \(AOD = 180 - 140 = 40\)

    Angle \(AOD = 40^\circ\)

    • (a) B1\(BOD = 140^\circ\)
    • (a) C1reason: vertically opposite angles are equal
    • (b) B1\(AOD = 40^\circ\)
    • (b) C1reason: angles on a straight line add up to 180 degrees
  8. Either paper 2 marks

    The diagram shows an isosceles triangle.

    x116°

    Work out the size of the angle marked \(x\).

    Show answer
    \(32^\circ\)

    The dashes show that two of the sides are equal, and in a triangle equal sides face equal angles. So the two base angles are the same size as each other, and \(x\) is one of them.

    All three angles add up to \(180^\circ\), so the two base angles share what is left after the \(116^\circ\):

    \(180 - 116 = 64\)

    \(x = 64 \div 2 = 32\)

    \(x = 32^\circ\)

    • M1\(180 - 116 = 64\) for the two base angles together
    • A1\(x = 32^\circ\)
  9. Either paper 2 marks

    Here is an isosceles triangle.

    59°x

    Work out the size of the angle marked \(x\).

    Show answer
    \(62^\circ\)

    The two dashed sides are equal, so the two angles facing them are equal. The angle marked \(59^\circ\) is one of that pair, so the angle at the other end of the unmarked side is \(59^\circ\) too.

    \(59 + 59 = 118\)

    The three angles of a triangle add up to \(180^\circ\), so

    \(x = 180 - 118 = 62\)

    \(x = 62^\circ\)

    • M1other base angle is \(59^\circ\) and the base angles total \(118\)
    • A1\(x = 62^\circ\)
  10. Either paper 3 marks

    Here is triangle \(ABC\). The side \(AB\) is continued to the point \(D\).

    ABC82°54°Dx
    (a)

    Work out the size of the angle marked \(x\).

    (b)

    A student says that an angle of \(156^\circ\) is an acute angle.

    The student is wrong. Explain why.

    Show answer
    (a) \(136^\circ\)
    (b) \(156^\circ\) is between 90° and 180°, so it is an obtuse angle. Acute angles are less than 90°.

    The angle marked \(x\) is an exterior angle of the triangle. It equals the sum of the two interior angles opposite it, which are the angles at \(A\) and \(C\).

    \(x = 82 + 54 = 136\)

    \(x = 136^\circ\)

    Check by the other route: the angle at \(B\) is \(180 - 82 - 54 = 44^\circ\), and angles on the straight line \(ABD\) add to \(180^\circ\), so \(x = 180 - 44 = 136^\circ\).

    Acute angles are less than 90°.

    \(156^\circ\) is between 90° and 180°, so it is not an acute angle. It is an obtuse angle.

    • (a) M1adds \(82\) and \(54\) to get \(136\), or finds the angle at B as \(44\)
    • (a) A1\(x = 136^\circ\)