Trigonometry: IGCSE Mathematics questions and answers
10 free IGCSE Mathematics Trigonometry questions written in the style of Cambridge Mathematics papers, each with its answer and mark scheme. 6 of them are A* challenge questions, built to catch the mistakes strong students still make.
This topic covers Pythagoras’ theorem, Right-angled triangles, Exact trigonometric values, Sine & cosine rules. Work through the questions below, then generate a full Trigonometry paper with a fresh set every time.
ABCDEFGH is a cuboid. The base ABCD is a horizontal rectangle with AB = 10 cm and BC = 5 cm. The vertical edges AE, BF, CG and DH are 5 cm long. M is the midpoint of AB.
(a)
Calculate the length of the diagonal AG.
[2]
(b)
Calculate the angle between AG and the base ABCD.
[2]
(c)
Calculate the angle between the line MG and the base ABCD.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
12.2 cm
AC² = 10² + 5², then AG² = AC² + 5² M1 12.2 A1
2
(b)
24.1°
tan = 5 ÷ AC M1 24.1 A1
2
(c)
35.3°
MC² = 5² + 5² M1 tan = 5 ÷ MC M1 35.3 A1
3
Question 2Structured · 4 marksA* challenge
The diagram shows a cuboid. It has a horizontal base measuring 5 cm by 8 cm and a height of 10 cm. Find the length of the longest diagonal of the cuboid and the angle it makes with the base.
[4]
Show answers and mark scheme
Part
Answer and marking points
Marks
13.7 cm; 46.7°
√(52 + 82 + 102) oe (base diagonal √(89) then 3D step) M1 13.7 cm A1 tan θ = 10 ÷ base diagonal M1 46.7° A1
4
Question 3Structured · 8 marksA* challenge
A boat sails 12 km from A to B on a bearing of 060°. It then sails 8 km from B to C on a bearing of 135°.
(a)
Show that angle ABC = 105°.
[2]
(b)
Calculate the distance AC.
[3]
(c)
Calculate the bearing on which the boat must sail to return directly from C to A.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
105°
uses co-interior / bearing of A from B = 240 M1 240 − 135 = 105 A1
sine rule for angle BAC M1 angle BAC = 28.8° A1 269° A1
3
Question 4Structured · 10 marksA* challenge
In triangle ABC, AB = 14 cm, AC = 10 cm and angle BAC = 69°.
Fig. 4.1
(a)
Calculate BC.
[3]
(b)
Calculate angle ABC.
[3]
(c)
Calculate the shortest distance from C to the line AB.
[2]
(d)
Calculate the area of triangle ABC.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
14 cm
BC2 = 102 + 142 − 2 × 10 × 14 × cos 69 M2 14 A1
3
(b)
41.9°
sin B10 = sin 6913.99M2 41.9 A1
3
(c)
9.34 cm
10 sin 69 M1 9.34 A1
2
(d)
65.4 cm²
½ × 10 × 14 × sin 69 M1 65.4 A1
2
Question 5Structured · 3 marks
The diagram shows triangle ABC.
Fig. 5.1
Calculate the length of AC.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
23.8 cm
ACsin 64 = 20sin 49M2 M1 for a correct implicit sine rule 23.8 A1
3
Question 6Structured · 3 marks
A ladder 6 m long leans against a vertical wall. The foot of the ladder is 1.7 m from the wall. How high up the wall does the ladder reach?
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
5.75 m
62 − 1.72M1 square root M1 5.75 A1
3
Question 7Structured · 2 marks
A rectangle measures 9 cm by 12 cm. Find the length of its diagonal.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
15 cm
√(92 + 122) M1 15 A1
2
Question 8Structured · 3 marks
(a)
Write down the exact value of cos 60°.
[1]
(b)
Write down the exact value of tan 30°.
[1]
(c)
Write down the exact value of cos 0°.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
12
12 oe B1
1
(b)
√33
√33 oe B1
1
(c)
1
1 oe B1
1
Question 9Structured · 6 marksA* challenge
The diagram shows a cuboid ABCDEFGH.
Fig. 9.1
(a)
Calculate the length of the diagonal AG.
[3]
(b)
Calculate the angle between AG and the base ABCD.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
13.4 cm
AC2 = 72 + 72M1 AG2 = 72 + 72 + 92M1 13.4 A1
3
(b)
42.3°
angle GAC identified B1 tan = 99.899 or sin = 913.38M1 42.3 A1
3
Question 10Structured · 10 marksA* challenge
From a point A on level ground, the angle of elevation of the top of a vertical tower is 30°. From a point B, 20 m closer to the tower on the same straight line, the angle of elevation is 45°.
(a)
Find the exact height of the tower, with a rational denominator.
[5]
(b)
A different tower of unknown height h stands at F. From P the elevation of its top is 30°, and from Q it is 45°. P and Q are 20 m apart on level ground and angle PFQ = 60°. Find h exactly.
[5]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
h = 20 ÷ (√3 − 1) = 10(√3 + 1)
BF = h (45°) B1 tan 30° = hh + 20 = 1√3M1 h√3 = h + 20 A1 h = 20√3 − 1M1 rationalised to 10(√3 + 1) A1
5
(b)
h = 20√(4 − √3) ≈ 13.3 m
FP = h√3, FQ = h B1 cosine rule in triangle PFQ M1 400 = 3h2 + h2 − 2 × h√3 × h × 12A1 h2(4 − √3) = 400 M1 h = 20√(4 − √3)A1
In the 0580 syllabus, Trigonometry covers Pythagoras’ theorem, Right-angled triangles, Exact trigonometric values, Sine & cosine rules.
Where can I get Trigonometry questions with mark schemes?
This page has 10 Trigonometry questions with answers. For more, the free generator on mentorriyaz.in builds a new Trigonometry paper with a full mark scheme each time you click.
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Riyaz Shaikh has mentored IGCSE, IB and A Level students for 15+ years, one-on-one, online and in Mumbai. The hardest questions on this site are the ones he uses to find a student's weak points.
Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in