Triangle Area Calculations Revision Notes | GCSE AQA Higher Maths

1. Why the Standard Area Formula Sometimes Fails

You already know that the area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$. That formula is reliable — but only when you actually know the perpendicular height. In many exam questions, and in real triangles, the height is never given. You are handed two sides and the angle squeezed between them, and nothing else.

This is where the formula $A = \frac{1}{2}ab\sin C$ steps in. It is the Higher-tier tool that unlocks the area of any triangle — acute, obtuse, or right-angled — directly from two sides and their included angle, with no need to find a perpendicular height first.

2. The Formula: $A = \frac{1}{2}ab\sin C$

The formula looks compact, but every letter carries a precise meaning. Getting the variable labels right is the first place marks are won or lost.

Notice that $C$ is the angle that sits between sides $a$ and $b$. If you pick two sides but use an angle that is not between them, the formula gives the wrong answer. Always check the angle is the one enclosed by your chosen sides.

A useful way to picture this: imagine sides $a$ and $b$ as two arms of a hinge. Angle $C$ is how wide the hinge is open. The wider the hinge (up to $90^\circ$), the larger $\sin C$ becomes, and so the larger the area. Beyond $90^\circ$, $\sin C$ starts to decrease again, but it stays positive for angles between $0^\circ$ and $180^\circ$, so the formula still gives a positive area for obtuse triangles. At exactly $90^\circ$, $\sin 90^\circ = 1$ and the formula reduces to $\frac{1}{2}ab$.

3. Calculating the Area — Worked Examples

The method follows a clean four-step pattern every time: identify the two sides and the included angle, write the formula, substitute, then calculate. Work through the example below step by step before trying the practice questions.

Worked example: Calculate the Area of a Triangle (calculator-allowed exam question)

Triangle $PQR$ has $PQ = 9$ cm, $PR = 6$ cm, and angle $P = 47^\circ$. Calculate the area of the triangle. Give your answer to 3 significant figures.

1. $a = PQ = 9$ cm, $b = PR = 6$ cm, $C = \angle P = 47^\circ$ — Angle $P$ sits between sides $PQ$ and $PR$, so it is the included angle. [M1 — correct identification]
2. $A = \frac{1}{2}ab\sin C$ — State the formula before substituting — this earns a method mark even if arithmetic slips later.
3. $A = \frac{1}{2} \times 9 \times 6 \times \sin 47^\circ$ — [M1 — correct substitution of all three values]
4. $\sin 47^\circ \approx 0.7314$ $A = \frac{1}{2} \times 9 \times 6 \times 0.7314$ $A = \frac{1}{2} \times 54 \times 0.7314$ $A = 27 \times 0.7314$ $A = 19.7478...$ — Keep the full decimal from your calculator — do not round until the final step.
5. $A \approx 19.7$ cm² — [A1 — correct answer to 3 significant figures with units]. Missing units costs a mark.

Answer: $A = 19.7$ cm² (to 3 s.f.)

Now try a slightly different version — this time the triangle is obtuse. Many students panic when the included angle is greater than $90^\circ$, but the formula works identically. Your calculator handles $\sin$ of obtuse angles without any adjustment needed from you.

Worked example: Area of a Triangle with an Obtuse Included Angle (calculator-allowed exam question)

Triangle $ABC$ has $AB = 12$ cm, $AC = 8$ cm, and angle $A = 130^\circ$. Calculate the area. Give your answer to 1 decimal place.

1. $a = AB = 12$ cm, $b = AC = 8$ cm, $C = \angle A = 130^\circ$ — Angle $A$ is between sides $AB$ and $AC$ — confirmed as the included angle.
2. $A = \frac{1}{2} \times 12 \times 8 \times \sin 130^\circ$ — [M1 — correct formula and substitution]
3. $\sin 130^\circ = \sin(180^\circ - 130^\circ) = \sin 50^\circ \approx 0.7660$ — Your calculator gives this directly. The note about $\sin(180^\circ - \theta)$ explains why obtuse angles still give a positive area — $\sin$ is positive in the second quadrant.
4. $A = \frac{1}{2} \times 12 \times 8 \times 0.7660$ $A = \frac{1}{2} \times 96 \times 0.7660$ $A = 48 \times 0.7660$ $A = 36.768$ — Carry the full decimal — rounding mid-calculation can shift the final answer.
5. $A \approx 36.8$ cm² — [A1 — correct to 1 decimal place with units]

Answer: $A = 36.8$ cm² (to 1 d.p.)

4. Finding a Missing Side or Angle Using the Area Formula

The specification requires you to calculate not only the area but also sides or angles of any triangle. When the area is given and one piece of information is missing, you rearrange $A = \frac{1}{2}ab\sin C$ to isolate the unknown — exactly as you would rearrange any other formula.

There are two rearrangement scenarios worth practising: finding a missing side when the area and the other side and angle are known, and finding a missing angle when the area and both sides are known. The second case requires an inverse sine at the end.

Worked example: Find a Missing Angle Given the Area and Two Sides (calculator-allowed exam question)

The area of triangle $DEF$ is $30$ cm². Sides $DE = 10$ cm and $DF = 8$ cm. Find angle $D$. Give your answer to 1 decimal place.

1. $A = \frac{1}{2}ab\sin C$ $30 = \frac{1}{2} \times 10 \times 8 \times \sin D$ — [M1 — correct substitution]. Angle $D$ is between sides $DE$ and $DF$, so it is the included angle.
2. $30 = \frac{1}{2} \times 80 \times \sin D$ $30 = 40\sin D$ — $\frac{1}{2} \times 10 \times 8 = \frac{1}{2} \times 80 = 40$
3. $\sin D = \frac{30}{40} = 0.75$ — [M1 — correct rearrangement to isolate $\sin D$]
4. $D = \sin^{-1}(0.75)$ $D = 48.5904...^\circ$ — Use the $\sin^{-1}$ (or arcsin) function on your calculator. Keep the full display value before rounding.
5. $D \approx 48.6^\circ$ — [A1 — correct to 1 decimal place]. Note: there is a second possible angle, $180^\circ - 48.6^\circ = 131.4^\circ$. Check whether this is geometrically valid given the triangle's other information.

Answer: $D = 48.6^\circ$ (to 1 d.p.)

The obtuse possibility in Step 5 is worth remembering. Because $\sin(180^\circ - \theta) = \sin \theta$, there are technically two angles between $0^\circ$ and $180^\circ$ that share the same sine value. In an exam, check whether the obtuse version is consistent with the triangle — if the other angles would then sum to more than $180^\circ$, discard it.

Worked example: Find a Missing Side Given the Area, One Side, and the Included Angle (calculator-allowed exam question)

Triangle $GHI$ has area $45$ cm², $GH = 15$ cm, and angle $G = 72^\circ$. Find the length $GI$. Give your answer to 3 significant figures.

1. $45 = \frac{1}{2} \times 15 \times GI \times \sin 72^\circ$ — [M1 — correct substitution]. $GI$ is the unknown side; angle $G$ is between $GH$ and $GI$.
2. $\sin 72^\circ \approx 0.9511$ — Read directly from your calculator.
3. $45 = \frac{1}{2} \times 15 \times 0.9511 \times GI$ $45 = 7.5 \times 0.9511 \times GI$ $45 = 7.1333... \times GI$ — $\frac{1}{2} \times 15 = 7.5$, then $7.5 \times 0.9511 = 7.1333...$
4. $GI = \frac{45}{7.1333...}$ $GI = 6.3089...$ — [M1 — correct rearrangement and division]
5. $GI \approx 6.31$ cm — [A1 — correct to 3 significant figures with units]

Answer: $GI = 6.31$ cm (to 3 s.f.)

5. Deliberate Practice

These questions follow the same structure as GCSE Higher exam questions. Work each one out fully before checking the answer — method marks depend on showing your substitution clearly.

6. Exam Technique and Common Mistakes

examiners report that the most common source of lost marks on this topic is not the formula itself — it is the details around it. The checklist below targets the errors that appear most frequently in mark schemes.

Bringing this all together: the formula $A = \frac{1}{2}ab\sin C$ is your go-to tool whenever a triangle question gives you two sides and the angle between them, whether the task is to find the area, a missing side, or a missing angle. Identify the included angle first, write the formula, substitute carefully, and round only at the end.

At GCSE, the formula $A = \frac{1}{2}ab\sin C$ is used in three directions: to find an area from two sides and the included angle, to rearrange for a missing side when the area is known, and to rearrange for a missing angle before applying inverse sine. The same formula works for acute, right, and obtuse triangles because $\sin C$ is positive for every angle between $0^\circ$ and $180^\circ$.