Pythagoras' theorem and trigonometry - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 36 exam-style questions on Pythagoras' theorem and trigonometry for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

36 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

The diagram shows a right-angled triangle ABC.

Work out the value of x.
Give your answer correct to 3 significant figures.

Geometry diagram, marked "Diagram NOT accurately drawn". Triangle ABC. Angle ACB is marked as a right angle (90°). Angle CAB is marked 34°. Side AB is labelled 12.5 cm. Side BC is labelled x cm.

Model answer

\(\sin 34^\circ = \dfrac{x}{12.5}\)
\(x = 12.5 \times \sin 34^\circ = 6.9899...\)
\(x = 6.99\) (3 s.f.)

Explanation

Label the sides from the 34° angle: 12.5 cm is the hypotenuse and x is opposite, so use sine (SOH). Make sure your calculator is in degrees. A common mistake is to use cosine, which gives 10.4.

Mark scheme

1 mark
M1 for \(\sin 34^\circ = \dfrac{x}{12.5}\) or \(12.5 \times \sin 34^\circ\)
oe, e.g. \(12.5 \times \cos 56^\circ\)
1 mark
A1 for 6.99
awrt 6.99 (6.989...); correct answer with no working scores full marks

Question 2

3 marks

The diagram shows a vertical building BT on horizontal ground.
Mia stands at G, 35 m from the foot of the building, B.
Her eyes, E, are 1.6 m above the ground.
The angle of elevation of the top of the building, T, from her eyes is 38°.

Work out the height of the building.
Give your answer correct to 3 significant figures.

Geometry diagram, marked "Diagram NOT accurately drawn". Line segments GB, BT, GE and ET are drawn. A dashed line is drawn from E going to the right. Angle GBT is marked as a right angle (90°). The angle at E between the line going to the right from E and ET is marked 38°. GB is labelled 35 m. GE is labelled 1.6 m.

Model answer

Height above Mia's eyes: \(35 \times \tan 38^\circ = 27.345...\) m
Height of building \(= 27.345... + 1.6 = 28.945...\) m
Height = 28.9 m (3 s.f.)

Explanation

The angle of elevation is measured from the horizontal at eye level, so the right-angled triangle has adjacent side 35 m and opposite side equal to the part of the building above eye level. Use tan, then add Mia's eye height. A common mistake is to forget the 1.6 m.

Mark scheme

1 mark
M1 for \(\tan 38^\circ = \dfrac{h}{35}\) or \(35 \times \tan 38^\circ\) (= 27.34...)
oe
1 mark
M1 for adding 1.6 to their \(35 \tan 38^\circ\)
1 mark
A1 for 28.9 m
awrt 28.9 (28.945...); correct answer with no working scores full marks

Question 3

3 marks

The diagram shows a circle with centre O and radius 7.8 cm.
A and B are points on the circle.
Angle AOB = 112°.

Work out the length of the chord AB.
Give your answer correct to 3 significant figures.

Geometry diagram, marked "Diagram NOT accurately drawn". A circle with centre O; A and B lie on it. Line segments OA, OB and AB are drawn. Angle AOB is marked 112°. OA is labelled 7.8 cm.

Model answer

Triangle AOB is isosceles (OA = OB = 7.8 cm). The perpendicular from O bisects AB and angle AOB.
Half of AB \(= 7.8 \times \sin 56^\circ = 6.4664...\) cm
\(AB = 2 \times 6.4664... = 12.932...\) cm
AB = 12.9 cm (3 s.f.)

Explanation

Two radii and a chord make an isosceles triangle. The line from the centre perpendicular to the chord bisects both the chord and the angle at the centre, giving right-angled triangles with hypotenuse 7.8 cm and angle 56°. A common mistake is to forget to double the half-chord.

Mark scheme

1 mark
M1 for splitting triangle AOB into two right-angled triangles with angle 56° at O, e.g. \(\sin 56^\circ = \dfrac{x}{7.8}\)
oe; x is half of AB
1 mark
M1 for \(2 \times 7.8 \times \sin 56^\circ\)
oe
1 mark
A1 for 12.9 cm
awrt 12.9 (12.932...); correct answer with no working scores full marks; use of the cosine rule is a fully correct alternative

Question 4

4 marks

A rectangular field is 85.6 m long and 47.3 m wide.
A farmer wants to put a straight fence along one diagonal of the field.

Fencing is sold in whole metres.
It costs £12.75 per metre.

Work out the cost of the fencing the farmer needs to buy.

Model answer

Length of diagonal \(= \sqrt{85.6^2 + 47.3^2} = \sqrt{9564.65} = 97.799...\) m
Fencing is sold in whole metres, so the farmer needs 98 m.
Cost \(= 98 \times 12.75 = £1249.50\)

Explanation

The diagonal is the hypotenuse of a right-angled triangle whose shorter sides are the length and width of the field. Because fencing is sold in whole metres, round up (not to the nearest metre) so there is enough. A common mistake is to work out the perimeter, or to multiply 97.799... by 12.75 without rounding up.

Mark scheme

1 mark
P1 for \(85.6^2 + 47.3^2\) (= 9564.65)
oe; correct answer with no working scores full marks
1 mark
A1 for 97.79... m
accept 97.7 to 97.8
1 mark
P1 for rounding up to 98 m and multiplying by 12.75
oe
1 mark
A1 for £1249.50
cao

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