IB DP Mathematics: Applications and Interpretation -- Geometry and Trigonometry Strand Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

Related: Geometry and Trigonometry study guide and revision notes.

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Section A

1. State which rule (sine or cosine) should be used when given three sides of a triangle and no angles. [1]

2. A hemisphere has radius $r$. State the formula for its curved surface area. [1]

3. State the two pieces of information needed to test whether two lines in three dimensions intersect, once their parametric equations are set equal. [2]

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Section B

4. Triangle $ABC$ has $a = 8$ cm, $b = 11$ cm, and the included angle $C = 52°$. Find side $c$, to 3 significant figures. [3]

5. Triangle $ABC$ has $\angle A = 40°$, $\angle B = 65°$, and side $a = 10$ cm (opposite $A$). Find side $b$, to 3 significant figures. [3]

6. A solid consists of a right circular cylinder of radius 3 cm and height 10 cm, topped with a solid hemisphere of the same radius.

(a) Find the total surface area of the solid, to 3 significant figures. [4] (b) Find the total volume of the solid, to 3 significant figures. [3]

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Section C

7. A ship sails from port A on a bearing of $070°$ for 40 km to point B, then changes course to a bearing of $150°$ and sails a further 25 km to point C.

(a) Find the interior angle $ABC$. [3] (b) Hence find the distance AC, to 3 significant figures. [3]

8. Line $L_1$ has vector equation $\mathbf{r} = \begin{pmatrix}1\2\0\end{pmatrix} + t\begin{pmatrix}1\-1\2\end{pmatrix}$ and line $L_2$ has vector equation $\mathbf{r} = \begin{pmatrix}3\0\4\end{pmatrix} + s\begin{pmatrix}-1\1\1\end{pmatrix}$.

(a) By setting the two equations equal, form three equations in $t$ and $s$. [2] (b) Determine whether the lines intersect, and if so, find the point of intersection. [4]

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Worked answers

1. The cosine rule (SSS gives no angle to start from directly, so the cosine rule is used first to find one angle). [1]

2. $2\pi r^2$. [1]

3. The values of the two parameters ($t$ and $s$, or equivalent) found from any two of the three component equations must also satisfy the third component equation – this consistency check is what confirms genuine intersection rather than the lines simply passing near each other in three dimensions. [2]

4. $c^2 = a^2 + b^2 - 2ab\cos C = 8^2 + 11^2 - 2(8)(11)\cos 52° = 64 + 121 - 176\cos 52° = 185 - 176(0.6157) = 185 - 108.36 = 76.64$. $c = \sqrt{76.64} = 8.755… \approx 8.76$ cm. [3]

5. By the sine rule, $\dfrac{b}{\sin B} = \dfrac{a}{\sin A}$, so $b = \dfrac{a \sin B}{\sin A} = \dfrac{10 \sin 65°}{\sin 40°} = \dfrac{10(0.9063)}{0.6428} = \dfrac{9.063}{0.6428} = 14.10…$ $\approx 14.1$ cm. [3]

6. (a) The solid’s outer surface consists of: the cylinder’s curved surface, $2\pi r h = 2\pi(3)(10) = 60\pi$; the cylinder’s flat base, $\pi r^2 = 9\pi$ (the top is fully covered by the hemisphere, so it is not part of the outer surface); and the hemisphere’s curved surface, $2\pi r^2 = 18\pi$. Total surface area $= 60\pi + 9\pi + 18\pi = 87\pi = 273.3… \approx 273$ cm². [4] (b) Cylinder volume $= \pi r^2 h = \pi(9)(10) = 90\pi$. Hemisphere volume $= \dfrac{2}{3}\pi r^3 = \dfrac{2}{3}\pi(27) = 18\pi$. Total volume $= 90\pi + 18\pi = 108\pi = 339.2… \approx 339$ cm³. [3]

7. (a) The ship’s bearing of travel into B was $070°$, so the reverse bearing from B back to A is $070° + 180° = 250°$. The ship’s onward bearing of travel from B is $150°$. The interior angle $ABC$ is the angle between direction BA ($250°$) and direction BC ($150°$): $250° - 150° = 100°$. [3] (b) $AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC) = 40^2 + 25^2 - 2(40)(25)\cos 100° = 1600 + 625 - 2000(-0.1736) = 2225 + 347.2 = 2572.2$. $AC = \sqrt{2572.2} = 50.72… \approx 50.7$ km. [3]

8. (a) Setting the equations equal component-by-component: $x$: $1 + t = 3 - s$, i.e. $t + s = 2$. $y$: $2 - t = s$, i.e. $t + s = 2$ (the same equation as from $x$, so not independent of it). $z$: $2t = 4 + s$, i.e. $2t - s = 4$. [2] (b) Using $t + s = 2$ and $2t - s = 4$: adding the two equations gives $3t = 6$, so $t = 2$, and then $s = 2 - 2 = 0$. Checking all three original components with $t = 2$, $s = 0$: $x$: $1 + 2 = 3$ and $3 - 0 = 3$ ✓; $y$: $2 - 2 = 0$ and $0 + 0 = 0$ ✓; $z$: $2(2) = 4$ and $4 + 0 = 4$ ✓. All three are consistent, so the lines do intersect, at the point $(3, 0, 4)$. [4]

A note on method

Question 7 is built to mirror the exact skill the study guide flags as the main source of lost marks in bearings problems: converting a bearing at the vertex where the ship changes course into a genuine interior triangle angle, rather than assuming the two given bearings can be subtracted directly without first accounting for the reverse bearing. A diagram sketched before any calculation begins – showing north lines at both A and B, the two bearings, and the resulting interior angle – is worth drawing even under time pressure, since it is far easier to spot a sign or direction error on a labelled sketch than in a line of algebra. Question 8 is a reminder that a system of three equations in two unknowns is, in general, overdetermined: two of the three component equations are used to solve for the parameters, and the third is then checked for consistency. If the third equation had not been satisfied, the correct conclusion would have been that the lines do not intersect (they are skew, assuming they are not parallel), not that an arithmetic error had necessarily been made – distinguishing “the lines genuinely don’t meet” from “I’ve made a mistake” is itself part of the skill this kind of HL question tests.

Official syllabus

International Baccalaureate Organization, Diploma Programme Subject Brief – Mathematics: Applications and Interpretation, first assessment 2021 – the same source cited by the study guide and revision notes. Verified 2026-09-06.