IB Maths AA SL · Unit 3: Geometry and Trigonometry
IB Maths AA SL Non Right Angled Trigonometry Questions
Exam-style IB Maths AA SL non 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.
- 29 questions
- Paper 1: 12
- Paper 2: 17
- 4 easy
- 12 medium
- 9 hard
- 2 very hard
- 2 starter
- 3 worked examples
Practise Non Right Angled Trigonometry questions →
AA SL formula booklet
What's examined in AA SL non right angled trigonometry
The question bank covers these non right angled trigonometry question types (number of questions in brackets):
- Cosine Rule calculations (13)
- Sine Rule & ambiguous case (10)
- Area of triangle (6)
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 →
Non Right Angled Trigonometry worked examples
Worked example 1: Cosine rule and triangle area · easy
In triangle $ABC$, $AB = 17$ m, $AC = 45$ m, included angle $B\hat{A}C = 38^\circ$. Calculate $BC$ and the area to 3 s.f.
1. Cosine Rule (SAS): $BC^2 = 17^2 + 45^2 - 2(17)(45)\cos 38^\circ$.
2. Evaluate: $BC^2 = 289 + 2025 - 1530(0.7880\ldots) = 1108.34\ldots$
3. Square root: $BC = \mathbf{33.3}$ m.
4. Area: $\frac{1}{2}(17)(45)\sin 38^\circ = 382.5 \times 0.6156\ldots = \mathbf{235\text{ m}^2}$.
Examiner tip: Check the angle mode on your calculator! It must be Degrees for this question.
Worked example 2: Angles of elevation and the sine rule · medium
From $M$ on flat ground, the angle of elevation to the top of a tree $Q$ is $20^\circ$. From $N$, which is $12$ m closer to the tree on the same horizontal line, the angle of elevation to $Q$ is $35^\circ$. Calculate the vertical height of the tree.
1. Obtuse angle at $N$ inside triangle $MNQ$: $180^\circ - 35^\circ = 145^\circ$.
2. Top angle: $180^\circ - 20^\circ - 145^\circ = 15^\circ$.
3. Sine Rule: $\frac{NQ}{\sin 20^\circ} = \frac{12}{\sin 15^\circ}$, so $NQ = 15.859\ldots$ m.
4. Right triangle: $\sin 35^\circ = \frac{\text{Height}}{15.859\ldots}$, giving Height $= \mathbf{9.10}$ m.
Examiner tip: Avoid rounding intermediate lengths like $NQ$. Keep the full unrounded value in the ANS memory to prevent premature rounding.
Worked example 3: The ambiguous case of the sine rule · hard
Triangle $ABC$ has $AB = 3\sqrt{2}$ cm, $BC = 3$ cm, angle $B\hat{A}C = 30^\circ$. Find the two possible exact values for angle $A\hat{C}B$ and explain why two triangles exist.
1. Sine Rule: $\frac{3\sqrt{2}}{\sin\hat{C}} = \frac{3}{\sin 30^\circ}$.
2. Rearrange: $\sin\hat{C} = \frac{3\sqrt{2}\sin 30^\circ}{3} = \frac{\sqrt{2}}{2}$.
3. Acute solution: $\hat{C}_1 = \mathbf{45^\circ}$.
4. Obtuse solution: $\hat{C}_2 = 180^\circ - 45^\circ = \mathbf{135^\circ}$.
5. Explain: two triangles exist because the side opposite the given angle ($BC = 3$) is strictly shorter than the adjacent side ($AB = 3\sqrt{2}$) but longer than the altitude $h = 3\sqrt{2}\sin 30^\circ \approx 2.12$.
Examiner tip: Sine Rule gives only the acute principal value. Manually check whether $180^\circ - \text{acute}$ also forms a valid triangle.
Try these IB Maths AA SL non 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 · 5 marks · Paper 2
Owen, Henry and Tom are rugby players passing a ball in a park. Owen is at point \(O\), Henry is at point \(H\) and Tom is at point \(T\). The distance between Owen and Henry is \(25\text{ m}\) and the distance between Henry and Tom is \(18\text{ m}\). The angle \(O\hat{H}T\) is \(96^\circ\).
Find the distance of the pass from Owen directly to Tom, \(OT\).
Find the size of the angle \(O\hat{T}H\).
Question 2 · medium · 6 marks · Paper 1
The following triangle shows triangle \(ABC\), with \(AB = 15\), \(AC = 20\), and \(BC = x\).
Given that \(\cos B\hat{A}C = \frac{2}{3}\):
Find the exact value of \(\sin B\hat{A}C\).
Find the exact area of triangle \(ABC\).
By finding the exact value of \(x\), show that triangle \(ABC\) is an isosceles triangle.
Question 3 · hard · 6 marks · Paper 2
A small airline operates between three locations \(A\), \(B\) and \(C\). \(B\) is located \(530\text{ km}\) from \(A\) on a bearing of \(248^\circ\). \(C\) is located \(300\text{ km}\) due East from the midpoint, \(M\), of the line segment \(AB\).
Find the size of the angle \(A\hat{M}C\).
Calculate the straight-line distance \(AC\).
All 29 non right angled trigonometry questions with mark schemes →
FAQ
How many IB Maths AA SL non right angled trigonometry questions are there?
There are 29 exam-style non right angled trigonometry questions in the AA SL question bank (Paper 1: 12 · Paper 2: 17), graded 4 easy, 12 medium, 9 hard, 2 very hard, 2 starter. Every question has a full IB-style mark scheme (M, A and R marks).
Is non right angled trigonometry on Paper 1 or Paper 2?
Both. In the bank, Paper 1: 12 · Paper 2: 17. 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.
Non Right Angled Trigonometry in other IB Maths courses
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