# The Sine Rule, Cosine Rule and Triangle Area

> Solving triangles without a right angle. Sine rule, cosine rule, area formula and the ambiguous case.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/sine-and-cosine-rules/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The lost hikers
Two rangers stand 5 km apart on a ridge in the Blue Mountains. Both can see the smoke from a lost hiker's fire. One measures the angle between the ridge and the smoke as $62^\circ$. The other measures $48^\circ$.

There is no right angle anywhere in this triangle. SOH CAH TOA will not work. But two angles and one side are enough to find exactly how far the smoke is from each ranger. The tool they use is the **sine rule**.

**In short:** The sine and cosine rules solve triangles that have no right angle. Use the sine rule $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$ when you know a side and its opposite angle, the cosine rule $c^2 = a^2 + b^2 - 2ab\cos C$ otherwise, and find the area with $\dfrac{1}{2}ab\sin C$.

## Starting point: Years 9 and 10

Label a triangle with capital letters $A$, $B$, $C$ for the angles and small letters $a$, $b$, $c$ for the sides. Side $a$ is **opposite** angle $A$, and so on.

*Diagram: Triangle with angles A, B, C and opposite sides a, b, c*
*Each side is named after the angle across from it.*

### The sine rule

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

Use it when you know a **side and its opposite angle**, plus one more side or angle.

### Worked example: The lost hikers
The third angle is $180^\circ - 62^\circ - 48^\circ = 70^\circ$, and it sits opposite the 5 km ridge. The distance $d$ from the first ranger is opposite the $48^\circ$ angle.

$$\frac{d}{\sin 48^\circ} = \frac{5}{\sin 70^\circ} \;\Rightarrow\; d = \frac{5 \sin 48^\circ}{\sin 70^\circ} \approx 3.95 \text{ km}$$

### Calculator tip: sin⁻¹ only gives acute angles
$\sin^{-1}$ always gives an angle between $-90^\circ$ and $90^\circ$. If the angle in the triangle could be obtuse, the other possible answer is $180^\circ$ minus the calculator's answer.

The cosine rule does not have this problem: $\cos^{-1}$ of a negative number gives the obtuse angle directly. Between steps, keep the full value with `Ans` or `STO` instead of a rounded one.

### The cosine rule

$$c^2 = a^2 + b^2 - 2ab\cos C$$

Use it when you know **two sides and the angle between them** (to find the third side), or **all three sides** (to find an angle). For an angle, rearrange:

$$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$

### Worked example: Third side
Two sides are 7 cm and 9 cm with $40^\circ$ between them.

$$c^2 = 49 + 81 - 2(7)(9)\cos 40^\circ \approx 33.48 \;\Rightarrow\; c \approx 5.79 \text{ cm}$$

### Area of any triangle

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

Two sides and the angle between them. For the triangle above, area $= \dfrac{1}{2}(7)(9)\sin 40^\circ \approx 20.2 \text{ cm}^2$.

### Key idea: Which rule?
- A matching side and angle pair is known: **sine rule**.
- Two sides and the included angle, or three sides: **cosine rule**.
- Two sides and the included angle, want area: $\dfrac{1}{2}ab\sin C$.

### Watch out: Square root at the end
After the cosine rule you have $c^2$. Remember to take the square root.

### Exam tip: The cosine rule contains Pythagoras
If $C = 90^\circ$, then $\cos C = 0$ and the rule becomes $c^2 = a^2 + b^2$.

**Mind map**

```text
Non-right triangles
  Labels
    Side a opposite angle A
  Sine rule
    a / sin A = b / sin B
    Needs a matching pair
  Cosine rule
    c² = a² + b² - 2ab cos C
    Two sides and included angle
    Three sides for an angle
  Area
    ½ ab sin C
  Ambiguous case
    Two possible triangles
```

### Year 11: Obtuse angles and the ambiguous case
Sine is positive for angles between $90^\circ$ and $180^\circ$ too: $\sin 150^\circ = \sin 30^\circ = \dfrac{1}{2}$. In general $\sin(180^\circ - \theta) = \sin \theta$.

So when the sine rule gives an **angle**, there may be two answers: $\theta$ and $180^\circ - \theta$. This is the **ambiguous case**.

**Example.** $a = 8$, $b = 10$, $A = 40^\circ$. Then $\sin B = \dfrac{10 \sin 40^\circ}{8} \approx 0.803$, so $B \approx 53.5^\circ$ or $B \approx 126.5^\circ$. Both work, because $40^\circ + 126.5^\circ < 180^\circ$. Two different triangles fit the data.

*Diagram: The ambiguous case. From A, side b of length 10 goes up at 40 degrees to C. A circle of radius 8 around C crosses the base line at two points, B1 and B2, so two triangles fit the data: one with an obtuse angle of 126.5 degrees at B1 and one with an acute angle of 53.5 degrees at B2*
*Side a = 8 can swing to two places on the base, shown by the dashed arc. B₁ gives an obtuse angle (126.5°) and B₂ an acute one (53.5°). They add to 180°, because sin(180° − θ) = sin θ.*

Cosine is negative for obtuse angles, so the cosine rule never gives two answers.

**Exact answers.** Use exact values in the rules. A triangle with sides 5 and 8 and the angle $60^\circ$ between them has area $\frac{1}{2}(5)(8)\sin 60^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}$. For an obtuse angle, $\sin 120^\circ = \frac{\sqrt{3}}{2}$, $\sin 150^\circ = \frac{1}{2}$ and $\cos 120^\circ = -\frac{1}{2}$.

*Diagram: A triangle with sides 5 and 8 and a 60 degree angle between them. Its height is 5 sin 60 degrees, so its area is 10 root 3*
*The height is the slanted side times the sine of the angle: 5 sin 60°. So the area is ½ × 8 × 5 sin 60° = 10√3.*

**Working back from an area.** If two equal sides $x$ meet at $150^\circ$ and the area is 16: $\frac{1}{2}x^2 \sin 150^\circ = \frac{x^2}{4} = 16$, so $x = 8$.

*Diagram: Two equal sides x meeting at 150 degrees. The height drops outside the triangle onto the base extended and equals x sin 150 degrees, which is half of x. The area is 16, so x is 8*
*With an obtuse angle the height drops outside the triangle, onto the base extended. It is still the slanted side times the sine of the angle: x sin 150° = x ÷ 2, so the area is ½ × x × x ÷ 2 = x² ÷ 4.*

### Year 11: Bearings with non-right triangles
A ship sails 12 km on a bearing of $050^\circ$, then 8 km on a bearing of $140^\circ$. The turn between the two legs is $90^\circ$ here, so Pythagoras gives the distance from the start: $\sqrt{144 + 64} \approx 14.4$ km. When the angle is not $90^\circ$, find the angle inside the triangle using parallel north lines, then use the cosine rule.

*Diagram: A ship sails 12 km on a bearing of 050 degrees, then 8 km on a bearing of 140 degrees. North lines are drawn at the start and at the turn. The turn is a right angle, so the distance from the start is 14.4 km*
*Draw a north line at every turn. The first leg points back on a bearing of 230° from the turn, and 230° − 140° = 90°, so the triangle has a right angle at the turn.*

**Worked example.** Two ships leave port together, one on a bearing of $040^\circ$ at 10 km/h and the other on $150^\circ$ at 12 km/h. After 2 hours they are 20 km and 24 km from port, and the angle between their paths is $150^\circ - 40^\circ = 110^\circ$. So the distance between them is
$$d^2 = 20^2 + 24^2 - 2(20)(24)\cos 110^\circ \approx 1304 \;\Rightarrow\; d \approx 36.1 \text{ km}$$

*Diagram: Two ships leave a port. One sails 20 km on a bearing of 040 degrees, the other 24 km on a bearing of 150 degrees. The angle between their paths is 110 degrees, and the distance between them is about 36.1 km*
*Both bearings are measured clockwise from the same north line, so the angle between the paths is 150° − 40° = 110°. Two sides and the angle between them: use the cosine rule.*

To find a bearing at the end, find the angle in the triangle with the sine rule, then add or subtract it from a north or south line.

### Year 12: The biggest triangle
Two sticks of length 6 and 10 are hinged at one end. The area of the triangle they make is $A = \dfrac{1}{2}(6)(10)\sin\theta = 30\sin\theta$.

The largest value of $\sin\theta$ is 1, at $\theta = 90^\circ$. So the biggest area is 30 square units. In Year 12, calculus confirms this: $\dfrac{dA}{d\theta} = 30\cos\theta = 0$ when $\theta = \dfrac{\pi}{2}$.

*Diagram: Two sticks hinged at O: one of length 10 along the ground and one of length 6. At 45 degrees and 135 degrees the height of the tip is 6 sin theta, about 4.2. At 90 degrees the height is the full 6, so the area is the biggest, 30*
*The base stays 10, so the area is ½ × 10 × height. The height is 6 sin θ, which is largest, 6, when the sticks are at right angles. At 45° and at 135° the height is the same, about 4.2, because sin 135° = sin 45°.*

## Quick summary

- Side $a$ is opposite angle $A$.
- Sine rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$. Needs a side with its opposite angle.
- Cosine rule: $c^2 = a^2 + b^2 - 2ab\cos C$. Two sides and the angle between, or three sides.
- Area $= \dfrac{1}{2}ab\sin C$.
- Finding an angle with the sine rule may give two answers: $\theta$ and $180^\circ - \theta$.

## Quiz

Try these questions from the chapter quiz.

1. In triangle $ABC$, side $b$ is opposite:
   - A. angle $A$
   - B. angle $B$
   - C. angle $C$
   - D. the right angle
   - Answer: B. angle $B$. Each side is named after the angle across from it.

2. Two angles of a triangle are $50^\circ$ and $70^\circ$. The third angle is:
   - A. $40^\circ$
   - B. $60^\circ$
   - C. $120^\circ$
   - D. $70^\circ$
   - Answer: B. $60^\circ$. Angles add to $180^\circ$.

3. You know two sides and the angle between them and want the third side. Use:
   - A. the sine rule
   - B. the cosine rule
   - C. SOH CAH TOA
   - D. the area formula
   - Answer: B. the cosine rule. Two sides and the included angle means cosine rule.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/sine-and-cosine-rules/#quiz