The Three Power Tools#
Before tackling 2D and 3D problems, you need three formulas at your fingertips. These work in any triangle — not just right-angled ones.
The Sine Rule#
$$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $$When to use it: When you have a side and its opposite angle as a pair, plus one more piece of information.
Typical scenarios:
- Given 2 angles and 1 side (AAS)
- Given 2 sides and an angle opposite one of them (SSA — the ambiguous case)
The Cosine Rule#
$$ a^2 = b^2 + c^2 - 2bc\cos A $$Or rearranged to find an angle:
$$ \cos A = \frac{b^2 + c^2 - a^2}{2bc} $$When to use it: When the Sine Rule doesn’t work — specifically:
- Given 3 sides (SSS) — find an angle
- Given 2 sides and the included angle (SAS) — find the third side
The Area Rule#
$$ \text{Area} = \frac{1}{2}ab\sin C $$When to use it: When you know two sides and the included angle and need the area.
Memory aid: The Sine Rule needs opposite pairs. The Cosine Rule and Area Rule need the included angle (the angle between the two known sides).
1. Choosing the Right Tool#
| Given | Unknown | Use |
|---|---|---|
| 2 angles + 1 side | Missing side | Sine Rule |
| 2 sides + opposite angle | Missing angle | Sine Rule |
| 3 sides | An angle | Cosine Rule |
| 2 sides + included angle | Third side | Cosine Rule |
| 2 sides + included angle | Area | Area Rule |
2. Worked 2D Examples#
Example 1: Sine Rule — Finding a Side#
In $\triangle PQR$: $\hat{P} = 40°$, $\hat{Q} = 75°$, $p = 12$ cm. Find $q$.
$\hat{R} = 180° - 40° - 75° = 65°$
$$ \frac{q}{\sin 75°} = \frac{12}{\sin 40°} $$$$ q = \frac{12 \sin 75°}{\sin 40°} = \frac{12 \times 0.9659}{0.6428} = 18.03 \text{ cm} $$Example 2: Cosine Rule — Finding a Side#
In $\triangle ABC$: $b = 8$, $c = 11$, $\hat{A} = 52°$. Find $a$.
$$ a^2 = 8^2 + 11^2 - 2(8)(11)\cos 52° $$$$ a^2 = 64 + 121 - 176 \times 0.6157 $$$$ a^2 = 185 - 108.36 = 76.64 $$$$ a = 8.75 \text{ cm} $$Example 3: Area Rule#
Find the area of $\triangle ABC$ with $b = 8$, $c = 11$, $\hat{A} = 52°$.
$$ \text{Area} = \frac{1}{2}(8)(11)\sin 52° = 44 \times 0.7880 = 34.67 \text{ cm}^2 $$3. The 3D Strategy: The Common Side#
3D Trigonometry is just 2D Trigonometry happening on different planes (usually one horizontal and one vertical).
The “Bridge” Analogy#
Imagine two islands (Triangle 1 and Triangle 2). They aren’t connected, but there is a bridge (the Common Side) that touches both.
- Use the information from one triangle to find the Common Side.
- Then use that side to solve the next triangle.
The 4-Step Method#
- Identify the planes: Which triangle is “flat” on the ground? Which one is “standing up”?
- Find the Common Side: The line segment that belongs to both triangles.
- Solve Triangle 1: Use the Sine Rule (or Cosine Rule) on the ground triangle to find the Common Side.
- Solve Triangle 2: Use the Common Side value in the vertical triangle to find the height, distance, or angle required.
Worked Example: Tower Problem#
A tower $TC$ stands vertically on level ground. From point $A$ on the ground, the angle of elevation to the top of the tower is $\alpha$. From point $B$, also on the ground, the angle of elevation is $\beta$. $AB = d$ and $\hat{ACB} = \theta$.
Prove that $h = \frac{d \sin\alpha \sin\beta}{\sin(\alpha - \beta)}$
Step 1: In $\triangle TAB$ (vertical plane containing both angles):
In $\triangle TCA$: $\tan\alpha = \frac{h}{AC}$, so $AC = \frac{h}{\tan\alpha}$
In $\triangle TCB$: $\tan\beta = \frac{h}{BC}$, so $BC = \frac{h}{\tan\beta}$
Step 2: In the triangle $\triangle TAB$, apply the Sine Rule on the vertical triangle.
Note: $\hat{TAB} = 90° - \alpha$ and $\hat{TBA} = 90° - \beta$, so $\hat{ATB} = \alpha - \beta$.
Using the Sine Rule in $\triangle TAB$:
$$ \frac{d}{\sin(\alpha - \beta)} = \frac{TB}{\sin(90° - \alpha)} = \frac{TB}{\cos\alpha} $$So: $TB = \frac{d\cos\alpha}{\sin(\alpha - \beta)}$
Step 3: In $\triangle TCB$: $\sin\beta = \frac{h}{TB}$
$$ h = TB \sin\beta = \frac{d\cos\alpha\sin\beta}{\sin(\alpha - \beta)} $$This can also be written as $h = \frac{d\sin\alpha\sin\beta}{\sin(\alpha - \beta)}$ depending on which triangle pair you choose.
4. Abstract Proofs#
Exams often ask you to “Prove that $h = \frac{d \sin \alpha \tan \theta}{\sin(\alpha + \beta)}$”.
Strategy:
- Treat $\alpha$, $\beta$, and $\theta$ exactly like you would treat $30°$ and $45°$.
- Express the Common Side in terms of $h$ using trig ratios in the vertical triangle.
- Find the Common Side using the Sine Rule in the ground triangle.
- Set the two expressions equal and solve for $h$.
- Use compound angle identities when you see $\sin(\alpha + \beta)$ in the answer.
🚨 Common Mistakes#
- Mixing up Angles of Elevation: The angle of elevation is always measured from the horizontal. Students often pick the wrong angle inside the vertical triangle.
- Assuming $90°$ on the ground: Just because a diagram looks “square” doesn’t mean the angles on the ground are $90°$. Unless you see the square box symbol, use the Sine/Cosine Rules.
- Wrong Sine Rule pairs: In the Sine Rule, the side must be opposite its paired angle. $\frac{a}{\sin A}$, NOT $\frac{a}{\sin B}$.
- Calculator in wrong mode: Ensure your calculator is in degrees mode, not radians.
- Not labelling the diagram: In 3D problems, labelling every angle and side (even with variables) prevents confusion about which triangle you’re working in.
💡 Pro Tip: Look for “Isosceles”#
3D problems often involve a point equidistant from two others (like a tower in the middle of a field). This creates isosceles triangles on the ground, meaning you have two equal angles and two equal sides for free!