Non-right triangles: Law of Sines, Law of Cosines, area

IB Mathematics Applications and Interpretation HL· 6 min read

1. The Law of Sines★★☆☆☆⏱ 15 min

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📘 Definition

Law of Sines

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

For any triangle with sides $a, b, c$ opposite angles $A, B, C$ respectively, the ratio of each side to the sine of its opposite angle is constant.

Example:

Used when you know two angles and one side, or two sides and a non-included angle (SSA).

📐 Worked Example

In triangle $A B C$ , $A = 4 0^{\circ}$ , $B = 6 0^{\circ}$ , $a = 8$ cm. Find the length of side $b$ .

1
Write the Law of Sines ratio for the known and unknown values:
2
$\frac{a}{\sin A} = \frac{b}{\sin B}$
3
Substitute the known values into the equation:
4
$\frac{8}{\sin 40^\circ} = \frac{b}{\sin 60^\circ}$
5
Rearrange to isolate $b$ :
6
$b = \frac{8 \times \sin 60^\circ}{\sin 40^\circ}$
7
Calculate with a calculator:
8
$b \approx 10.8 \text{ cm}$

2. The Law of Cosines★★☆☆☆⏱ 15 min

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📘 Definition

Law of Cosines

c^2 = a^2 + b^2 - 2ab\cos C \quad \text{or} \quad \cos C = \frac{a^2 + b^2 - c^2}{2ab}

Relates the three sides of a triangle to one of its angles. It is a general case of Pythagoras' theorem, which only applies to right triangles.

Example:

Used when you know three sides, or two sides and the included angle between them.

📐 Worked Example

In triangle $A B C$ , sides $a = 5$ cm, $b = 7$ cm, included angle $C = 4 5^{\circ}$ . Find the length of side $c$ .

1
Select the form for finding an unknown side:
2
$c^2 = a^2 + b^2 - 2ab\cos C$
3
Substitute the known values:
4
$c^2 = 5^2 + 7^2 - 2(5)(7)\cos 45^\circ$
5
Calculate each term:
6
$c^2 = 25 + 49 - 70(0.7071) \approx 24.50$
7
Take the positive square root (side length cannot be negative):
8
$c \approx \sqrt{24.50} \approx 4.95 \text{ cm}$

3. Area of a Non-Right Triangle★★☆☆☆⏱ 10 min

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📘 Definition

Area of Any Triangle

\text{Area} = \frac{1}{2}ab\sin C

The area of any triangle is half the product of two sides multiplied by the sine of the included angle between them.

Example:

Works for all triangles, right or non-right, when two sides and the included angle are known.

📐 Worked Example

Find the area of triangle $A B C$ with $a = 10$ m, $b = 14$ m, included angle $C = 3 0^{\circ}$ .

1
Substitute into the area formula:
2
$\text{Area} = \frac{1}{2}(10)(14)\sin 30^\circ$
3
Simplify using $sin 3 0^{\circ} = 0.5$ :
4
$\text{Area} = 70 \times 0.5 = 35 \text{ m}^2$

If you know all three sides, first use the Law of Cosines to find any included angle, then apply this area formula. If you know two angles and one side, find the third angle, use the Law of Sines to find a second side, then calculate area.

4. Choosing the Right Rule & The Ambiguous Case★★★★☆⏱ 20 min

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Knowing which rule to use saves time on exams and avoids unnecessary calculation. Use this guide:

Given information	Recommended rule
Two angles + any side	Law of Sines
Two sides + non-included angle (SSA)	Law of Sines (check ambiguous case)
Two sides + included angle (SAS)	Law of Cosines + Area formula
Three sides (SSS)	Law of Cosines

For SSA (two sides, non-included angle), there can be 0, 1, or 2 valid triangles. This is the ambiguous case, a common exam topic.

📐 Worked Example

Given $a = 7$ , $b = 9$ , $A = 2 5^{\circ}$ , how many valid triangles exist?

1
Use Law of Sines to solve for $sin B$ :
2
$\sin B = \frac{b \sin A}{a} = \frac{9 \sin 25^\circ}{7} \approx 0.543$
3
Two angles between $0^{\circ}$ and $18 0^{\circ}$ have this sine value:
4
$B_1 \approx 32.9^\circ, \quad B_2 = 180^\circ - 32.9^\circ = 147.1^\circ$
5
Check if $B_{2}$ forms a valid triangle:
6
$147.1^\circ + 25^\circ = 172.1^\circ < 180^\circ$
7
Conclusion: Both angles are valid, so 2 distinct triangles exist.

5. Common Pitfalls

Wrong move:

Using Law of Sines for two sides and an included angle (SAS)

Why:

Law of Sines requires the known angle to be opposite a known side, which is not true for SAS

Correct move:

Use Law of Cosines for all SAS problems

Wrong move:

Forgetting to check the second angle in ambiguous SSA cases

Why:

IB exams explicitly test recognition of two valid triangles, so you will lose marks for missing the second solution

Correct move:

Always calculate the supplementary angle and check if it forms a valid triangle

Wrong move:

Using radian mode on a calculator for triangle problems

Why:

All IB triangle problems use degrees unless stated otherwise, leading to drastically wrong results

Correct move:

Confirm your calculator is in degree mode before starting any trigonometry problem

Wrong move:

Using the non-included angle in the $\frac{1}{2} ab sin C$ area formula

Why:

The formula only works when the angle is between the two sides you use

Correct move:

Confirm your angle is included between the two sides, or find the included angle first with Law of Sines/Cosines

Wrong move:

Keeping the negative square root when solving for a side with Law of Cosines

Why:

Side length is always positive, so including the negative root is an error

Correct move:

Only take the positive square root when calculating side length

6. Quick Reference Cheatsheet

Rule	Formula	When to use
Law of Sines	$\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C}$	2 angles + 1 side, SSA
Law of Cosines (side)	$c^{2} = a^{2} + b^{2} - 2 ab cos C$	SAS, find third side
Law of Cosines (angle)	$cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}$	SSS, find angle
Area of Triangle	$\frac{1}{2} ab sin C$	SAS, find area

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · Paper 1
Find unknown side via Law of Cosines
2024 · Paper 2
Calculate area of non-right triangle
2023 · Paper 1
Solve ambiguous SSA case with Law of Sines

Going deeper

Formula sheetIB AI HL Trigonometry Formula ReferenceOfficial syllabus formula page

What's Next

Mastering non-right triangle trigonometry is foundational for many higher topics in IB AI HL, including 2D navigation with bearings, and modelling real-world scenarios involving distances and angles. The rules you learned here extend directly to problems involving triangles embedded in 3D shapes, where you will break complex 3D problems into multiple 2D non-right triangles to solve for unknown heights or distances. These methods also underpin work with area of compound shapes and trigonometric modelling, which appear regularly on both Paper 1 and Paper 2 exams.