Trigonometry: past paper questions

Real questions on trigonometry

The newest 20 of 69 written questions. Open one to see it on the real paper.

1. 0980/21 · October/November 2025 · Q26[3]
   26 Solve $\tan x=-\frac{1}{\sqrt{3}}$ for $0^{\circ }⩽x⩽360^{\circ }$.See it on the paper →
2. 0980/41 · October/November 2025 · Q13[3]
   13 [Diagram: Two vertical poles $AE$ and $PT$ standing on horizontal ground $AP$. $AE=1.5\text{ m}$, $PT=12.6\text{ m}$, $AP=33\text{ m}$. Labelled "NOT TO SCALE".] The diagram shows two vertical poles, $AE$ and $PT$, standing on horizontal ground, $AP$. Calculate the angle of elevation of the point $T$ from the...See it on the paper →
3. 0980/41 · October/November 2025 · Q21[2]
   21 In triangle $STU$, $ST=8$ cm, $SU=9$ cm and angle $TSU=50^{\circ }$. Calculate the area of triangle $STU$. .................................................... cm${}^2$See it on the paper →
4. 0980/41 · October/November 2025 · Q29a[3]
   [Diagram: A quadrilateral $ABCD$ divided into two triangles by diagonal $AC$. Triangle $ABC$ has side $AB=10\text{ cm}$, side $BC=13\text{ cm}$, and side $AC=14\text{ cm}$. The angle $\angle BAC$ is labelled $y^{\circ }$. Triangle $ADC$ has angle $\angle ACD=38^{\circ }$ and angle $\angle ADC=97^{\circ }$. The diagram...See it on the paper →
5. 0980/41 · October/November 2025 · Q29b[5]
   [Diagram: A quadrilateral $ABCD$ with diagonal $AC$. Triangle $ABC$ has side $AB=10\text{ cm}$, side $BC=13\text{ cm}$, and side $AC=14\text{ cm}$. The angle $\angle BAC$ is labelled $y^{\circ }$. The angle $\angle BCA$ is labelled $38^{\circ }$. Triangle $ADC$ has angle $\angle ADC=97^{\circ }$. The diagram is marked "...See it on the paper →
6. 0980/32 · May/June 2025 · Q24[5]
   24 The diagram shows two right-angled triangles. [Diagram: A composite figure consisting of two right-angled triangles, $△ABD$ and $△DBC$, sharing a common side $BD$. In $△ABD$, the right angle is at $D$. Side $AB$ is labeled $14.7\text{ m}$ and side $AD$ is labeled $10.8\text{ m}$. In $\tria...See it on the paper →
7. 0980/42 · May/June 2025 · Q20a[2]
   [Diagram: Quadrilateral $ABCD$ with diagonal $DB$. $CD=10$ m, $DB=12$ m. Angle $CDB=56^{\circ }$. Angle $ADB=34^{\circ }$. Angle $DBA=90^{\circ }$.] The diagram shows a quadrilateral $ABCD$. $CD=10$ m and $DB=12$ m. Angle $DBA=90^{\circ }$, angle $CDB=56^{\circ }$ and angle $ADB=34^{\circ }$. (a) Calculate the lengt...See it on the paper →
8. 0980/42 · May/June 2025 · Q20b[3]
   [Diagram: A quadrilateral $ABCD$ with diagonal $DB$. $CD=10$ m, $DB=12$ m. Angle $CDB=56^{\circ }$. Angle $ADB=34^{\circ }$. Angle $DBA=90^{\circ }$. NOT TO SCALE] The diagram shows a quadrilateral $ABCD$. $CD=10$ m and $DB=12$ m. Angle $DBA=90^{\circ }$, angle $CDB=56^{\circ }$ and angle $ADB=34^{\circ }$. (b) Calcu...See it on the paper →
9. 0980/42 · May/June 2025 · Q20c[5]
   [Diagram: A quadrilateral $ABCD$ with diagonal $DB$. $CD=10\text{ m}$, $DB=12\text{ m}$. Angle $CDB=56^{\circ }$. Angle $ADB=34^{\circ }$. Angle $DBA=90^{\circ }$.] The diagram shows a quadrilateral $ABCD$. $CD=10\text{ m}$ and $DB=12\text{ m}$. Angle $DBA=90^{\circ }$, angle $CDB=56^{\circ }$ and angle $ADB = 34^\c...See it on the paper →
10. 0980/42 · May/June 2025 · Q20d[3]
    [Diagram: A quadrilateral $ABCD$ with vertices $A,B,C,D$. Side $CD=10\text{ m}$. Diagonal $DB=12\text{ m}$. Angle $CDB=56^{\circ }$. Angle $ADB=34^{\circ }$. Angle $DBA=90^{\circ }$.] The diagram shows a quadrilateral $ABCD$. $CD=10\text{ m}$ and $DB=12\text{ m}$. Angle $DBA=90^{\circ }$, angle $CDB=56^{\circ }$...See it on the paper →
11. 0980/21 · October/November 2024 · Q20[5]
    [Diagram: A geometric diagram showing two right-angled triangles, $PQS$ and $RQT$, sharing a vertical line segment $QS$. Point $Q$ is at the bottom of the vertical line, with a right-angle symbol between $PQ$ and $QS$. Point $P$ is to the left of $Q$ on a horizontal line. Point $R$ is to the right of $Q$ on the same ho...See it on the paper →
12. 0980/21 · October/November 2024 · Q21[3]
    21 Solve the equation $3\tan x+5=1$ for $0^{\circ }⩽x⩽360^{\circ }$.See it on the paper →
13. 0980/41 · October/November 2024 · Q6a[5]
    The diagram shows a field $ABCD$. A straight path $AC$ goes across the field. [Diagram: Quadrilateral $ABCD$ with diagonal $AC$. $AD=420\text{ m}$, $DC=830\text{ m}$, $AB=1150\text{ m}$. Angle $ADC=106^{\circ }$. Angle $ABC=62^{\circ }$. Label "NOT TO SCALE".] (a) Show that $AC=1028\text{ m}$, correct to the near...See it on the paper →
14. 0980/41 · October/November 2024 · Q6b[4]
    The diagram shows a field $ABCD$. A straight path $AC$ goes across the field. [Diagram: Quadrilateral $ABCD$ with diagonal $AC$. $AD=420\text{ m}$, $DC=830\text{ m}$, $AB=1150\text{ m}$. Angle $ADC=106^{\circ }$. Angle $ABC=62^{\circ }$. Label "NOT TO SCALE".] (b) Angle $ACB$ is obtuse. Calculate angle $ACB$.See it on the paper →
15. 0980/41 · October/November 2024 · Q6c[4]
    The diagram shows a field $ABCD$. A straight path $AC$ goes across the field. [Diagram: Quadrilateral $ABCD$ with diagonal $AC$. $AD=420\text{ m}$, $DC=830\text{ m}$, $AB=1150\text{ m}$. Angle $ADC=106^{\circ }$, Angle $ABC=62^{\circ }$. Label "NOT TO SCALE".] (c) Part of the field, triangle $ACD$, is sold for $41\...See it on the paper →
16. 0980/41 · October/November 2024 · Q10b(i)[4]
    [Diagram: A sector OAB of a circle with centre O. The radius OA is labelled $r$ cm. The angle AOB is $50^{\circ }$. The region between the chord AB and the arc AB is shaded. The diagram is marked NOT TO SCALE.] The diagram shows a sector $OAB$ of a circle, with centre $O$, and a chord $AB$. The shaded segment has area $30...See it on the paper →
17. 0980/41 · October/November 2024 · Q10b(ii)[4]

    4-mark “Calculate” question — its maths and diagrams are best seen on the paper.

    See it on the paper →
18. 0980/12 · May/June 2024 · Q18[2]
    18 [Diagram: A right-angled triangle. The hypotenuse is labelled 14 cm. The base is labelled 8.5 cm. The angle between the hypotenuse and the base is labelled $x^{\circ }$. The right angle is at the bottom left. The text "NOT TO SCALE" is present.] The diagram shows a right-angled triangle. Calculate the value of $x$.See it on the paper →
19. 0980/12 · May/June 2024 · Q25a[4]
    [Diagram: A triangle $ADC$ with a vertical line segment $BD$ dropping from vertex $D$ to the base $AC$. $A$, $B$, and $C$ are collinear. The angle at $B$ (angle $ABD$) is marked as a right angle. The angle at $C$ (angle $BCD$) is marked as $48^{\circ }$. The length of side $AD$ is labeled $21.2\text{ m}$. The length of se...See it on the paper →
20. 0980/12 · May/June 2024 · Q25b[3]
    The diagram shows a flagpole, $BD$, held by two ropes, $AD$ and $CD$. $ABC$ is a straight line and angle $ABD=90^{\circ }$. $AD=21.2\text{ m}$, $AB=16.5\text{ m}$ and angle $BCD=48^{\circ }$. [Diagram: A triangle $ADC$ with a vertical line $BD$ from $D$ to $B$ on the base $AC$. Angle $ABD$ is a right angle. Angle $BC...See it on the paper →

What examiners said

This question was a challenge for many candidates. Most of those who attempted it understood that it required the use of Pythagoras’ theorem and trigonometry. Most realised the need to find length BD. Some premature rounding of BD to 9.9 or 10 led to answers outside the acceptable range. Inaccuracy also arose when candidates chose a longer route, finding CD first then using CD to find d. Some cand...

— Examiner report, May/June 2025, 0980/32 Q24 (Cambridge)

Most candidates were able to use a correct method to find the length of AB, the most efficient being AB sin(34°) sin(56°) tan(34°) = . Some used = or other longer methods. Some candidates made 12 AB 12 the error of giving their answer to 2 or fewer significant figures, i.e. 8.1 or 8, and not showing a more accurate value in their working, so could not be awarded the accuracy mark.

— Examiner report, May/June 2025, 0980/42 Q20(a) (Cambridge)

Many candidates were able to use a correct method to find the area of the quadrilateral ABCD. Some candidates lost accuracy because they used 8 or 8.1 for the length of AB or because they used a rounded value for sin(56°) within the method. Some common incorrect methods seen were, 1 1 for example, area of triangle BDC = 10 12 instead of 10 12 sin(56°), and area of 2 2 1 1 triangle ADB = 12 their A...

— Examiner report, May/June 2025, 0980/42 Q20(b) (Cambridge)