IB Math AA SL Topic 3.2: Sine & cosine rules | Notes & Flashcards | Aimnova

IB Math AA SL — Sine & cosine rules

Topic 3.2 of IB Mathematics: Analysis and Approaches covers Sine & cosine rules, which is part of Unit 3: Geometry & Trigonometry. Students explore key concepts including Right-angled trig, Sine & cosine rules, Area of a triangle. A strong understanding of sine & cosine rules is essential for IB Math AA SL exams and builds the foundation for connected topics across the syllabus.

Key concepts in Sine & cosine rules

Key Idea: This topic is your toolkit for solving any triangle — finding missing sides, angles and areas. It runs through geometry and trig questions on both papers, almost always worked by hand on Paper 1 (the GDC only does the final arithmetic).

📐 Right-angled triangles: SOH-CAH-TOA

\sin\theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos\theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan\theta = \frac{\text{opp}}{\text{adj}}

$\text{hyp}$: hypotenuse — longest side, opposite the right angle
$\text{opp}$: side opposite the angle θ you are using
$\text{adj}$: side next to θ (not the hypotenuse)

Know an angle + a side → form the ratio and rearrange for the missing side. Know two sides → form the ratio and take the inverse (sin⁻¹, cos⁻¹, tan⁻¹) for the angle. Three sides, no angle? Use Pythagoras a² + b² = c² — trig is only for when an angle is involved.

🔺 Any triangle: the sine & cosine rules

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

$a$: side opposite angle A (same for b/B, c/C)
$\frac{\sin A}{a} = \frac{\sin B}{b}$: flip it the other way up when you want an angle

a^2 = b^2 + c^2 - 2bc\cos A \qquad \cos A = \frac{b^2 + c^2 - a^2}{2bc}

$A$: angle opposite side a — must match the side on the left
$b,\ c$: the two sides enclosing angle A

📏 Area of a triangle

\text{Area} = \tfrac{1}{2}\,ab\,\sin C

$a,\ b$: the two sides you know
$C$: the **included** angle — the one *between* sides a and b

Important: The angle in ½ab·sinC must be the included angle (between the two sides you use), not just any angle. If the included angle is missing, find it first with the cosine rule (SSS) or the sine rule, then apply the area formula.

✏️ IB-style worked examples

IB-style question — right-angled triangle (Paper 1)

A ladder leans against a wall, reaching 6 m up. The ladder makes an angle of 65° with the ground. Find the length of the ladder.

Step by step:

The 6 m is opposite the 65° angle; the ladder is the hypotenuse → use sin.
$\sin 65^\circ = \frac{6}{L}$
Rearrange (unknown is on the bottom): multiply up, divide.
$L = \frac{6}{\sin 65^\circ} \approx 6.62$

Final answer:

The ladder is ≈ 6.62 m long.

IB-style question — sine & cosine rules (Paper 1)

In triangle PQR, p = 9, r = 11 and the included angle Q = 58°. Find side q, then angle P.

Step by step:

Two sides + the angle between (SAS) → cosine rule for q.
$q^2 = 9^2 + 11^2 - 2(9)(11)\cos 58^\circ$
Evaluate.
$q^2 = 202 - 104.9\ldots \Rightarrow q \approx 9.85$
Now there's a pair (q opposite Q) → sine rule, flipped, for P.
$\frac{\sin P}{9} = \frac{\sin 58^\circ}{9.85}$
Solve.
$\sin P = \frac{9\sin 58^\circ}{9.85} \Rightarrow P \approx 50.7^\circ$

Final answer:

q ≈ 9.85, P ≈ 50.7°.

IB-style question — area, then a missing side (Paper 1)

A triangular garden has two sides of 8 m and 11 m with an included angle of 40°. Find its area, then the length of the third side.

Step by step:

Included angle given → area straight away.
$\text{Area} = \tfrac{1}{2}(8)(11)\sin 40^\circ \approx 28.3$
Third side: SAS again → cosine rule.
$x^2 = 8^2 + 11^2 - 2(8)(11)\cos 40^\circ$
Evaluate.
$x^2 = 185 - 134.8\ldots \Rightarrow x \approx 7.09$

Final answer:

Area ≈ 28.3 m²; third side ≈ 7.09 m.

Important: When you find an angle from a sine value, there are two answers: the acute one and its obtuse partner (180° − it). Your GDC only shows the acute one. e.g. sin C = 0.5 gives C = 30° or 150°. Use the context (the longest side faces the largest angle; or whether the angle is labelled acute/obtuse) to pick the right one — or keep both. cos⁻¹ is safe (one answer in a triangle); sin⁻¹ is the trap.

Tap each card to reveal the answer.

Exam Tips

First decide the rule: right angle → SOH-CAH-TOA; SAS/SSS → cosine; side-with-opposite-angle → sine.
In the cosine rule, the side on the left must be opposite the angle in the cos term.
Finding an angle from sin? Check the obtuse partner 180° − θ — the GDC hides it.
Area needs the INCLUDED angle (between the two sides); find it first if it's missing.
Keep full accuracy in working and round only the final answer to 3 s.f.

IB Math AA SL — Sine & cosine rules

Key concepts in Sine & cosine rules

Key Idea: This topic is your toolkit for solving any triangle — finding missing sides, angles and areas. It runs through geometry and trig questions on both papers, almost always worked by hand on Paper 1 (the GDC only does the final arithmetic).

📐 Right-angled triangles: SOH-CAH-TOA

\sin\theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos\theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan\theta = \frac{\text{opp}}{\text{adj}}

$\text{hyp}$: hypotenuse — longest side, opposite the right angle
$\text{opp}$: side opposite the angle θ you are using
$\text{adj}$: side next to θ (not the hypotenuse)

Know an angle + a side → form the ratio and rearrange for the missing side. Know two sides → form the ratio and take the inverse (sin⁻¹, cos⁻¹, tan⁻¹) for the angle. Three sides, no angle? Use Pythagoras a² + b² = c² — trig is only for when an angle is involved.

🔺 Any triangle: the sine & cosine rules

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

$a$: side opposite angle A (same for b/B, c/C)
$\frac{\sin A}{a} = \frac{\sin B}{b}$: flip it the other way up when you want an angle

a^2 = b^2 + c^2 - 2bc\cos A \qquad \cos A = \frac{b^2 + c^2 - a^2}{2bc}

$A$: angle opposite side a — must match the side on the left
$b,\ c$: the two sides enclosing angle A

📏 Area of a triangle

\text{Area} = \tfrac{1}{2}\,ab\,\sin C

$a,\ b$: the two sides you know
$C$: the **included** angle — the one *between* sides a and b

Important: The angle in ½ab·sinC must be the included angle (between the two sides you use), not just any angle. If the included angle is missing, find it first with the cosine rule (SSS) or the sine rule, then apply the area formula.

✏️ IB-style worked examples

IB-style question — right-angled triangle (Paper 1)

A ladder leans against a wall, reaching 6 m up. The ladder makes an angle of 65° with the ground. Find the length of the ladder.

Step by step:

The 6 m is opposite the 65° angle; the ladder is the hypotenuse → use sin.
$\sin 65^\circ = \frac{6}{L}$
Rearrange (unknown is on the bottom): multiply up, divide.
$L = \frac{6}{\sin 65^\circ} \approx 6.62$

Final answer:

The ladder is ≈ 6.62 m long.

IB-style question — sine & cosine rules (Paper 1)

In triangle PQR, p = 9, r = 11 and the included angle Q = 58°. Find side q, then angle P.

Step by step:

Two sides + the angle between (SAS) → cosine rule for q.
$q^2 = 9^2 + 11^2 - 2(9)(11)\cos 58^\circ$
Evaluate.
$q^2 = 202 - 104.9\ldots \Rightarrow q \approx 9.85$
Now there's a pair (q opposite Q) → sine rule, flipped, for P.
$\frac{\sin P}{9} = \frac{\sin 58^\circ}{9.85}$
Solve.
$\sin P = \frac{9\sin 58^\circ}{9.85} \Rightarrow P \approx 50.7^\circ$

Final answer:

q ≈ 9.85, P ≈ 50.7°.

IB-style question — area, then a missing side (Paper 1)

A triangular garden has two sides of 8 m and 11 m with an included angle of 40°. Find its area, then the length of the third side.

Step by step:

Included angle given → area straight away.
$\text{Area} = \tfrac{1}{2}(8)(11)\sin 40^\circ \approx 28.3$
Third side: SAS again → cosine rule.
$x^2 = 8^2 + 11^2 - 2(8)(11)\cos 40^\circ$
Evaluate.
$x^2 = 185 - 134.8\ldots \Rightarrow x \approx 7.09$

Final answer:

Area ≈ 28.3 m²; third side ≈ 7.09 m.

Important: When you find an angle from a sine value, there are two answers: the acute one and its obtuse partner (180° − it). Your GDC only shows the acute one. e.g. sin C = 0.5 gives C = 30° or 150°. Use the context (the longest side faces the largest angle; or whether the angle is labelled acute/obtuse) to pick the right one — or keep both. cos⁻¹ is safe (one answer in a triangle); sin⁻¹ is the trap.

Tap each card to reveal the answer.

Exam Tips

First decide the rule: right angle → SOH-CAH-TOA; SAS/SSS → cosine; side-with-opposite-angle → sine.
In the cosine rule, the side on the left must be opposite the angle in the cos term.
Finding an angle from sin? Check the obtuse partner 180° − θ — the GDC hides it.
Area needs the INCLUDED angle (between the two sides); find it first if it's missing.
Keep full accuracy in working and round only the final answer to 3 s.f.

IB Math AA SL — Sine & cosine rules

Key concepts in Sine & cosine rules

📐 Right-angled triangles: SOH-CAH-TOA

🔺 Any triangle: the sine & cosine rules

📏 Area of a triangle

✏️ IB-style worked examples

IB-style question — right-angled triangle (Paper 1)

IB-style question — sine & cosine rules (Paper 1)

IB-style question — area, then a missing side (Paper 1)

Exam Tips

What you'll learn in Topic 3.2

Study resources — 3.2 Sine & cosine rules

Right-angled trig

Sine & cosine rules

Area of a triangle

Ready to study Sine & cosine rules?

Ready to practice?

IB Math AA SL — Sine & cosine rules

Key concepts in Sine & cosine rules

📐 Right-angled triangles: SOH-CAH-TOA

🔺 Any triangle: the sine & cosine rules

📏 Area of a triangle

✏️ IB-style worked examples

IB-style question — right-angled triangle (Paper 1)

IB-style question — sine & cosine rules (Paper 1)

IB-style question — area, then a missing side (Paper 1)

Exam Tips

What you'll learn in Topic 3.2

Study resources — 3.2 Sine & cosine rules

Right-angled trig

Sine & cosine rules

Area of a triangle

Ready to study Sine & cosine rules?

Ready to practice?