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.

Use $\sin^2 A + \cos^2 A = 1$ : $\sin B\hat{A}C = \sqrt{1 - \left(\frac{2}{3}\right)^2} = \sqrt{\frac{5}{9}}$ . (M1) $\sin B\hat{A}C = \frac{\sqrt{5}}{3}$ . A1 Area $= \frac{1}{2}(15)(20)\sin B\hat{A}C = 150 \times \frac{\sqrt{5}}{3}$ . (M1) Area $= 50\sqrt{5}$ . A1 Use the cosine rule: $x^2 = 15^2 + 20^2 - 2(15)(20)\left(\frac{2}{3}\right)$ . M1 $x^2 = 625 - 400 = 225 \implies x = 15$ . A1 Since $AB = BC = 15$ , two sides are equal, so $\triangle ABC$ is isosceles. AG

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IB Maths AA SL · Unit 3: Geometry and Trigonometry

Sine Rule, Cosine Rule & Area (IB Maths AA SL): notes, exam questions and worked examples

Q: How do I know whether to use the sine rule or the cosine rule?

If you know a side and the angle opposite it, use the sine rule. If you know two sides and the angle between them, or all three sides, use the cosine rule.

Q: Are the sine and cosine rules in the IB formula booklet?

Yes. The sine rule, both forms of the cosine rule and the area formula ½ab sin C are all in the AA SL booklet. The condition for the ambiguous case is not, so learn when two triangles are possible.

Q: How many IB Maths AA SL non right angled trigonometry questions are there?

There are 45 exam-style non right angled trigonometry questions in the AA SL question bank (Paper 1: 19 · Paper 2: 26), graded 8 easy, 19 medium, 16 hard, 2 very hard. Every question has a full IB-style mark scheme (M, A and R marks).

Q: Is non right angled trigonometry on Paper 1 or Paper 2?

Both. In the bank, Paper 1: 19 · Paper 2: 26. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.

Updated October 2026 · By Pete Bromfield

Use the sine rule when you know a side and its opposite angle, the cosine rule when you know two sides and the angle between them (or all three sides), and $A = \frac{1}{2} a b \sin C$ for the area. All three are in the formula booklet. Watch for the ambiguous case of the sine rule, where two different triangles fit the data.

45 questions
Paper 1: 19
Paper 2: 26
3 worked examples

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Notes
Formulas
Worked examples
Questions
FAQ

Sine Rule, Cosine Rule & Area notes

The sine rule

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ , with each side opposite its angle. You need one complete pair (a side and the angle opposite it) plus one more side or angle. Flip it to $\frac{\sin A}{a} = \frac{\sin B}{b}$ when finding an angle.

The cosine rule

$c^{2} = a^{2} + b^{2} - 2 a b \cos C$ finds a side from two sides and the included angle; $\cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$ finds an angle from three sides. With $a = 5$ , $b = 7$ and $C = 60^{\circ}$ : $c^{2} = 39$ , so $c = \sqrt{39} \approx 6.24$ .

Area of a triangle

$A = \frac{1}{2} a b \sin C$ , using two sides and the angle between them. For the triangle above, $A = \frac{1}{2} (5) (7) \sin 60^{\circ} = \frac{35 \sqrt{3}}{4} \approx 15.2$ .

The ambiguous case

Given two sides and a non-included acute angle, there can be two triangles: $\sin B = k$ has an acute answer $B$ and an obtuse answer $180^{\circ} - B$ . With $A = 30^{\circ}$ , $b = 10$ and $a = 6$ : $\sin B = \frac{5}{6}$ , so $B \approx {56.4}^{\circ}$ or $B \approx {123.6}^{\circ}$ , and both fit because the angles still add to less than $180^{\circ}$ .

Bearings, elevation and 3D

Draw a clear diagram first. Bearings are measured clockwise from north; angles of elevation and depression are measured from the horizontal. In 3D, find the right-angled or non-right-angled triangle in a plane and work in that triangle. Set your calculator to degrees for these questions.

Key formulas

Sine rule In the booklet: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
Cosine rule In the booklet: $c^{2} = a^{2} + b^{2} - 2 a b \cos C, \cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$
Area of a triangle In the booklet: $A = \frac{1}{2} a b \sin C$
Ambiguous case: given $a, b, A$ with $A$ acute, two triangles if $b \sin A < a < b$ Learn it

The IB formula booklet is provided in every exam paper. In the booklet = printed there, so know where to find it; Learn it = not in the booklet, so learn it by heart. All AA SL formulas →

Explained, with a worked example: Area of a triangle · Cosine rule · Sine rule

Sine Rule, Cosine Rule & Area worked examples

Worked example 1: Cosine rule and triangle area · easy

In triangle $A B C$ , $A B = 17$ m, $A C = 45$ m, included angle $B \hat{A} C = 38^{\circ}$ . Calculate $B C$ and the area to 3 s.f.

Solution

4 steps

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.

Solution

4 steps

Examiner tip: Avoid rounding intermediate lengths like

N Q

. 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 $A B C$ has $A B = 3 \sqrt{2}$ cm, $B C = 3$ 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.

Solution

5 steps

Examiner tip: Sine Rule gives only the acute principal value. Manually check whether

180^{\circ} - acute

also forms a valid triangle.

Sine Rule, Cosine Rule & Area exam-style questions

3 questions from our AA SL bank, written in the style of IB papers, easiest first. One has its full worked solution open here. Open the worked answer, or check your answer 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 $25 m$ and the distance between Henry and Tom is $18 m$ . The angle $O \hat{H} T$ is $96^{\circ}$ .

Find the distance of the pass from Owen directly to Tom, $O T$ .
Find the size of the angle $O \hat{T} H$ .

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Question 2 · medium · 6 marks · Paper 1

The following triangle shows triangle $A B C$ , with $A B = 15$ , $A C = 20$ , and $B C = x$ . 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 $A B C$ .
By finding the exact value of $x$ , show that triangle $A B C$ is an isosceles triangle.

Full worked solution Free, no sign-in

Use $\sin^{2} A + \cos^{2} A = 1$ : $\sin B \hat{A} C = \sqrt{1 - {(\frac{2}{3})}^{2}} = \sqrt{\frac{5}{9}}$ . (M1)
$\sin B \hat{A} C = \frac{\sqrt{5}}{3}$ . A1
Area $= \frac{1}{2} (15) (20) \sin B \hat{A} C = 150 \times \frac{\sqrt{5}}{3}$ . (M1)
Area $= 50 \sqrt{5}$ . A1
Use the cosine rule: $x^{2} = 15^{2} + 20^{2} - 2 (15) (20) (\frac{2}{3})$ . M1
$x^{2} = 625 - 400 = 225 ⟹ x = 15$ . A1
Since $A B = B C = 15$ , two sides are equal, so $△ A B C$ is isosceles. AG

Every question here has a full worked solution. See them all →

Question 3 · hard · 6 marks · Paper 2

A small airline operates between three locations $A$ , $B$ and $C$ . $B$ is located $530 km$ from $A$ on a bearing of $248^{\circ}$ . $C$ is located $300 km$ due East from the midpoint, $M$ , of the line segment $A B$ .

Find the size of the angle $A \hat{M} C$ .
Calculate the straight-line distance $A C$ .

Attempt it and see the mark scheme →

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Sine Rule, Cosine Rule & Area FAQ

How do I know whether to use the sine rule or the cosine rule?

If you know a side and the angle opposite it, use the sine rule. If you know two sides and the angle between them, or all three sides, use the cosine rule.

Are the sine and cosine rules in the IB formula booklet?

Yes. The sine rule, both forms of the cosine rule and the area formula ½ab sin C are all in the AA SL booklet. The condition for the ambiguous case is not, so learn when two triangles are possible.

How many IB Maths AA SL non right angled trigonometry questions are there?

There are 45 exam-style non right angled trigonometry questions in the AA SL question bank (Paper 1: 19 · Paper 2: 26), graded 8 easy, 19 medium, 16 hard, 2 very hard. 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: 19 · Paper 2: 26. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.

Next steps

More AA SL Unit 3: Geometry and Right Angled Trigonometry · Radians · Solving Trigonometric Equations · The Unit Circle and Exact Values of Trigonometric Ratios · Trigonometric Functions · Trigonometric Identities

Non Right Angled Trigonometry in other courses: AI SL

Teaching this topic? Free ready-to-teach lessons (slides and teacher notes): Sine rule · Cosine rule · Area = ½ ab sin C.

Keep revising AA SL: AA SL revision notes · IB Maths AA SL past papers · AA SL common exam mistakes · AA SL syllabus checklist · AA SL geometry and trigonometry knowledge organiser

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Sine Rule, Cosine Rule & Area (IB Maths AA SL): notes, exam questions and worked examples

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Sine Rule, Cosine Rule & Area notes

The sine rule

The cosine rule

Area of a triangle

The ambiguous case

Bearings, elevation and 3D

Key formulas

Sine Rule, Cosine Rule & Area worked examples

Worked example 1: Cosine rule and triangle area · easy

Solution

Worked example 2: Angles of elevation and the sine rule · medium

Solution

Worked example 3: The ambiguous case of the sine rule · hard

Solution

Sine Rule, Cosine Rule & Area exam-style questions

Question 1 · easy · 5 marks · Paper 2

Question 2 · medium · 6 marks · Paper 1

Full worked solution Free, no sign-in

Question 3 · hard · 6 marks · Paper 2

Sine Rule, Cosine Rule & Area FAQ

Next steps

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