Year 10 Mathematics · Measurement and Geometry · 20m read

Advanced Trigonometry

Ever wondered how surveyors measure land without perfect right angles, or how planes navigate using triangles? The sine and cosine rules let you solve any triangle, not just right-angled ones — they're the toolkit professionals use daily. You'll learn to find missing sides and angles in real situations, from mapping terrain to engineering, skills that appear constantly in VCE maths and beyond.

Learning objective: Apply the sine rule, cosine rule, and area formula to non-right-angled triangles.

By the end of this lesson you can:

I use the sine rule for angles and sides.
I use the cosine rule for sides and angles.
I calculate triangle area using the sine formula.

The unit circle is a Ferris wheel. As a seat goes round, its height traces the sine graph and its sideways position traces cosine. Trig is just tracking a Ferris wheel seat.

Sine rule, cosine rule, and area

EXTENDING TRIGONOMETRY BEYOND RIGHT TRIANGLES

For any triangle with sides $a$, $b$, $c$ and opposite angles $A$, $B$, $C$:

### THE SINE RULE
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$

Or equivalently (for finding angles):
$$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$$

Use the Sine Rule when you know:
• Two angles and one side (AAS or ASA) → find any other side.
• Two sides and a non-included angle (SSA) → find other angles (AMBIGUOUS CASE!).

### THE AMBIGUOUS CASE (SSA)
When given two sides and a NON-included angle, there may be:
• 0 solutions (impossible triangle)
• 1 solution (unique triangle)
• 2 solutions (two valid triangles!)

Check: if $\sin B > 1$ from the formula → impossible. If $0 < \sin B < 1$, consider whether $B$ is acute OR obtuse.

### THE COSINE RULE
$$a^2 = b^2 + c^2 - 2bc\cos A$$

Rearranged to find an angle:
$$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$

Use the Cosine Rule when you know:
• Three sides (SSS) → find any angle.
• Two sides and the INCLUDED angle (SAS) → find the opposite side.

Note: When $A = 90°$, $\cos A = 0$, and the formula reduces to Pythagoras! $a^2 = b^2 + c^2$.

### AREA OF A TRIANGLE
$$\text{Area} = \frac{1}{2}ab\sin C$$

Use when two sides and the INCLUDED angle are known.

### Choosing the Right Rule

| Given information | Use |
|------------------|-----|
| Right-angled triangle | SOH CAH TOA, Pythagoras |
| Two angles + one side | Sine rule |
| Two sides + non-included angle | Sine rule (check ambiguous case!) |
| Two sides + included angle | Cosine rule (for third side) OR Area formula |
| Three sides | Cosine rule (for any angle) |

### Bearings (compass directions)
• Bearings are measured CLOCKWISE from North.
• 000° = North, 090° = East, 180° = South, 270° = West.
• Used in navigation problems with sine/cosine rules.

### Australian context
Surveyors in Western Australia's mining regions use the sine and cosine rules to calculate distances across terrain that is too difficult to measure directly. GPS relies on similar trigonometric principles.

Sine Rule, Cosine Rule, Area

Sine rule: \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}
Cosine rule (side): a^2 = b^2 + c^2 - 2bc\cos A
Cosine rule (angle): \cos A = \dfrac{b^2 + c^2 - a^2}{2bc}
Area: \text{Area} = \dfrac{1}{2}ab\sin C

Rule selection

Sine rule: Two angles + one side, OR two sides + non-included angle.
Cosine rule: Two sides + included angle, OR three sides.
Area: $\frac{1}{2}ab\sin C$ — needs two sides and the angle BETWEEN them.

Worked example (Medium)

In triangle $ABC$, $A = 50°$, $B = 70°$, $a = 8$ cm. Find side $b$.

Sine rule. $\dfrac{b}{\sin B} = \dfrac{a}{\sin A}$
Substitute. $\dfrac{b}{\sin 70°} = \dfrac{8}{\sin 50°}$
Solve. $b = \dfrac{8 \sin 70°}{\sin 50°} = \dfrac{8 \times 0.9397}{0.7660} \approx 9.81$ cm

Answer: $b \approx 9.81$ cm.

Worked example (Hard)

Two ships leave a port. Ship A travels 12 km on a bearing of 060°. Ship B travels 8 km on a bearing of 140°. Find the distance between the ships.

Angle at port. Angle between the two directions: $140° - 60° = 80°$.
Cosine rule. $c^2 = 12^2 + 8^2 - 2(12)(8)\cos 80°$
Calculate. $c^2 = 144 + 64 - 192 \times 0.1736 = 208 - 33.33 = 174.67$
Distance. $c = \sqrt{174.67} \approx 13.22$ km

Answer: The ships are approximately 13.2 km apart.

Common mistakes

Using the sine rule when the cosine rule is needed (given two sides + included angle) The sine rule requires you to know a side opposite a KNOWN angle. If the angle is between the two known sides (included), you need the cosine rule. Fix: Check: do you know an angle that is OPPOSITE one of the given sides? Yes → sine rule. No (included angle) → cosine rule.
Forgetting the ambiguous case in SSA problems When you use the sine rule to find an angle and $0 < \sin \theta < 1$, there are potentially TWO angles ($\theta$ and $180° - \theta$). Both may give valid triangles. Fix: After finding the acute angle, check if the obtuse angle ($180° - \theta$) is also valid: the three angles must sum to 180° and all be positive.
Squaring brackets incorrectly in the cosine rule — computing $2bc\cos A$ incorrectly The cosine rule formula has a subtraction: $a^2 = b^2 + c^2 - 2bc\cos A$. Forgetting to subtract gives too large an answer. Fix: Write out the formula clearly: $a^2 = b^2 + c^2 - 2bc\cos A$. Substitute values carefully. Use brackets around $2bc\cos A$.

Practice questions

To find the third side of a triangle given two sides and the INCLUDED angle, use:
1. Sine rule
2. Cosine rule
3. SOH CAH TOA
4. Area formula
(1 marks)
A triangle has sides $b = 7$, $c = 5$, and $A = 60°$. Find side $a$.
(2 marks)
Find the area of triangle $PQR$ where $p = 9$ cm, $q = 12$ cm, and $R = 35°$.
(3 marks)
In triangle $ABC$, $a = 6$, $b = 8$, $c = 10$. Find angle $C$ using the cosine rule.
(3 marks)

Going Deeper — 3D Trigonometry

Trigonometry extends into THREE DIMENSIONS, combining multiple triangles in a single problem.

Key strategy: Identify the 2D triangles within the 3D figure, then solve step by step.

Classic problem: A vertical pole $AB$ stands at corner $A$ of a rectangular paddock $ABCD$ (50 m × 30 m). The angle of elevation to the top of the pole from corner $C$ is 20°. Find the height of the pole.

Step 1: Find distance $AC$ (diagonal of rectangle).
$AC = \sqrt{50^2 + 30^2} = \sqrt{3400} \approx 58.31$ m

Step 2: Use right triangle with $AC$ as base:
$\tan 20° = h / AC \Rightarrow h = 58.31 \times \tan 20° \approx 21.23$ m

Another type: Finding the angle between a diagonal and the base of a rectangular box.
For a box $l \times w \times h$:
Space diagonal $= \sqrt{l^2 + w^2 + h^2}$
Angle with base $= \tan^{-1}\!\left(\dfrac{h}{\sqrt{l^2 + w^2}}\right)$

3D problems appear regularly in Victorian Certificate examinations.

Summary

Sine rule: $a/\sin A = b/\sin B = c/\sin C$ — use for AAS, ASA, SSA.
Cosine rule: $a^2 = b^2 + c^2 - 2bc\cos A$ — use for SAS, SSS.
Area: $\frac{1}{2}ab\sin C$ — needs two sides and included angle.
Ambiguous case (SSA): check if supplement angle also gives a valid triangle.
Always label the triangle diagram with all given values before choosing a rule.