IB Maths AA SL · Unit 3: Geometry and Trigonometry
IB Maths AA SL Geometry and Right Angled Trigonometry Questions
Exam-style IB Maths AA SL geometry and right angled trigonometry questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step the way an IB examiner expects — then try the practice questions and check your working against the mark scheme.
- 34 questions
- Paper 1: 19
- Paper 2: 15
- 6 easy
- 12 medium
- 10 hard
- 4 very hard
- 2 starter
- 3 worked examples
Practise Geometry and Right Angled Trigonometry questions →
AA SL formula booklet
What's examined in AA SL geometry and right angled trigonometry
The question bank covers these geometry and right angled trigonometry question types (number of questions in brackets):
- Lengths and Angles in 3D (16)
- Geometric Mensuration and Ratios (13)
- 3D Coordinate Geometry (5)
Key formulas
- Double-angle: cosine
- \(\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta\)
- Double-angle: sine
- \(\sin 2\theta = 2\sin\theta\cos\theta\)
- Tangent from sine & cosine
- \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\)
- Cosine rule
- \(c^2 = a^2 + b^2 - 2ab\cos C\)
- Sine rule
- \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)
- Area of a triangle
- \(A = \tfrac{1}{2}ab\sin C\)
- Pythagorean identity
- \(\sin^2\theta + \cos^2\theta = 1\)
In the same notation as the IB formula booklet. All AA SL formulas →
Geometry and Right Angled Trigonometry worked examples
Worked example 1: Distance and midpoint in 3D · easy
Points $A(1, -2, 3)$ and $B(5, 4, -1)$ are located in a 3D Cartesian coordinate system. Find the exact distance between $A$ and $B$, and the exact coordinates of their midpoint $M$.
1. Identify the 3D distance formula: $AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
2. Substitute: $AB = \sqrt{(5-1)^2 + (4-(-2))^2 + (-1-3)^2}$.
3. Simplify: $AB = \sqrt{16+36+16} = \sqrt{68} = \mathbf{2\sqrt{17}}$.
4. Midpoint formula: $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$.
5. Substitute and evaluate: $\mathbf{M(3, 1, 1)}$.
Examiner tip: When calculating distance, always use brackets around negative numbers before squaring in your GDC; typing $-4^2$ instead of $(-4)^2$ gives $-16$ instead of $+16$.
Worked example 2: Slant height and surface area of a cone · medium
A right circular cone has a base radius of $r$ and a perpendicular height of $2\sqrt{2}r$. Find an exact expression for the slant height $l$ in terms of $r$, and hence the exact ratio of the cone's curved surface area to its flat base area.
1. Apply Pythagoras: $l^2 = r^2 + (2\sqrt{2}r)^2$.
2. Expand: $l^2 = r^2 + 8r^2 = 9r^2$.
3. Square root: $\mathbf{l = 3r}$.
4. Formulas: CSA $= \pi r l$, base $= \pi r^2$.
5. Substitute: CSA $= \pi r(3r) = 3\pi r^2$.
6. Ratio: $3\pi r^2 : \pi r^2 = \mathbf{3:1}$.
Examiner tip: When squaring surd expressions like $2\sqrt{2}r$, the square applies to every factor: $(2\sqrt{2}r)^2 = 4 \cdot 2 \cdot r^2 = 8r^2$.
Worked example 3: Angle between a 3D diagonal and a face · hard
A cube has side length $a$. Prove that the exact angle $\theta$ between its main interior diagonal and its square base satisfies $\sin\theta = \frac{1}{\sqrt{3}}$.
1. Base diagonal via Pythagoras: $\sqrt{a^2+a^2} = a\sqrt{2}$.
2. Main diagonal via Pythagoras again: $\sqrt{(a\sqrt{2})^2 + a^2} = \sqrt{3a^2} = a\sqrt{3}$.
3. Sine ratio: opposite is the vertical edge $a$, hypotenuse is $a\sqrt{3}$.
4. Simplify: $\sin\theta = \frac{a}{a\sqrt{3}} = \mathbf{\frac{1}{\sqrt{3}}}$.
Examiner tip: Always draw a separate flat 2D right-angled triangle for the cross-section, labelling "Opposite" and "Adjacent" before applying SOH CAH TOA.
Try these IB Maths AA SL geometry and right angled trigonometry questions
Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area.
Question 1 · easy · 3 marks · Paper 1
Find the exact length of the line segment joining the 3D coordinates $A(1, 2, 3)$ and $B(4, -2, 5)$.
Question 2 · medium · 6 marks · Paper 1
A solid object is constructed by placing a solid hemisphere of radius \(r\) precisely on top of a solid cylinder of radius \(r\) and height \(h\).
Given that the volume of the hemisphere is equal to the volume of the cylinder, express \(h\) in terms of \(r\).
Show that the total surface area of the combined solid is exactly \(\frac{13}{3}\pi r^2\).
Question 3 · hard · 6 marks · Paper 2
A model building is created in the shape of a rectangular-based right pyramid \(ABCDE\). The base \(ABCD\) is a rectangle with \(AB = 4.6\text{ cm}\) and \(BC = 7.2\text{ cm}\). The apex \(E\) is directly above the centre of the base, and each of the four slant edges \([AE]\), \([BE]\), \([CE]\) and \([DE]\) has length \(8.3\text{ cm}\).
Calculate the vertical height of the pyramid.
Calculate the volume of the model.
Find the angle that the triangular face \(ABE\) (with base \(4.6\text{ cm}\)) makes with the rectangular base \(ABCD\).
All 34 geometry and right angled trigonometry questions with mark schemes →
FAQ
How many IB Maths AA SL geometry and right angled trigonometry questions are there?
There are 34 exam-style geometry and right angled trigonometry questions in the AA SL question bank (Paper 1: 19 · Paper 2: 15), graded 6 easy, 12 medium, 10 hard, 4 very hard, 2 starter. Every question has a full IB-style mark scheme (M, A and R marks).
Is geometry and right angled trigonometry on Paper 1 or Paper 2?
Both. In the bank, Paper 1: 19 · Paper 2: 15. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.
Where can I get the mark schemes?
Open the AA SL Unit 3 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.
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