Year 10 Maths Advanced Trigonometry

Area Rule

Q: What is the area rule for a triangle?

The area of a triangle is one half times two sides times the sine of the angle between them: A = one half a b sin C. It works for any triangle, not just right-angled ones.

Q: Does the area rule work for obtuse angles?

Yes. The sine of an obtuse angle is positive and equals the sine of its supplement, so sin 130 degrees equals sin 50 degrees. Substitute the obtuse angle directly.

20 practice questions 1 video lesson Theory + worked examples

The area rule finds the area of any triangle from two sides and the angle between them: $A=\dfrac{1}{2}ab\sin C$. In NSW Year 10 Advanced maths it is used for exact areas with special angles, worked backwards to find a side or an angle, and combined with the sine and cosine rules (NESA outcome MA5‑TRG‑P‑01).

You will find areas of triangles and quadrilaterals, find a side or angle from a given area, and use the cosine rule first when only three sides are known. Practical problems include topsoil for a garden bed, a boat sail and a regular octagon inside a circle. You will find both angles that give the same area and explain why there are two, find the area of an equilateral triangle from its side, and find the largest possible area of a triangle. It is part of Year 10 Advanced Trigonometry.

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Theory

The area rule, $A = \dfrac{1}{2}ab\sin C$, gives the area of any triangle from two sides and the angle between them, with no height needed. This Year 10 Advanced maths page covers exact areas, working backwards to a side or an angle, triangles given by three sides, and areas of quadrilaterals and regular polygons. It comes up if you are heading for Advanced or Extension maths in Years 11 and 12.

The area rule says the area of a triangle is half the product of two sides times the sine of the included angle, the angle between those two sides.

It comes from $A = \dfrac{1}{2} \times \text{base} \times \text{height}$: with base $a$, the height is $b\sin C$. It works for acute and obtuse angles, because $\sin(180^\circ - C) = \sin C$.

Since $\sin C \le 1$, the largest area two sides can make is $\dfrac{1}{2}ab$, when the angle between them is $90^\circ$.

Working backwards from an area gives $\sin C$. In a triangle that usually means two possible angles, $C$ and $180^\circ - C$, and both make a triangle with the same area.

The height from $A$ is $h = b\sin C$, so area $= \dfrac{1}{2} \times a \times b\sin C$. The angle must be the one between the two sides you use.

A regular polygon with $n$ sides splits into $n$ identical triangles with two radii and an angle of $\dfrac{360^\circ}{n}$ at the centre.

\[A = \dfrac{1}{2}ab\sin C = \dfrac{1}{2}bc\sin A = \dfrac{1}{2}ac\sin B\]

A = \frac{1}{2} a b \sin C

\[\text{Equilateral triangle with side } s\text{:}\quad A = \dfrac{\sqrt{3}}{4}s^2\]

A = \frac{\sqrt{3}}{4} s^{2}

Exact values. \(\sin 30^\circ = \sin 150^\circ = \dfrac{1}{2}\), \(\sin 45^\circ = \sin 135^\circ = \dfrac{1}{\sqrt{2}}\), \(\sin 60^\circ = \sin 120^\circ = \dfrac{\sqrt{3}}{2}\). Use these for exact areas, and to solve for a side exactly.

Which rule finds the angle you need? Two sides and the included angle: area rule straight away. Three sides: cosine rule for an angle first. Two angles and a side: sine rule for a second side first.

How to use the area rule

Identify two sides and the angle between them. If the angle you know is not between them, find another side or angle first.
Substitute into $A = \dfrac{1}{2}ab\sin C$. Use exact values if an exact answer is asked for.
Answer with square units, rounded as asked. Use the unrounded area for costs.

Working backwards and composite shapes

Unknown side: substitute the area and solve the linear equation. If the sides are $x$ and $x + 4$, you get a quadratic: expand, factorise and reject negative lengths.
Unknown angle: solve for $\sin C$, then give $C$ and $180^\circ - C$ unless the diagram shows which one.
Quadrilaterals: split along a diagonal into two triangles. Find the diagonal with the cosine rule if you need it, then add the two areas. A parallelogram is two congruent triangles.
Regular polygons: join the centre to each vertex and multiply the area of one triangle by $n$ (Example 4).
Largest area: the area $\dfrac{1}{2}ab\sin\theta$ is greatest when $\sin\theta = 1$, that is $\theta = 90^\circ$.

Example 1 — Area and cost

A triangular park has two sides of $46$ m and $38$ m meeting at an angle of $74^\circ 15'$.
(i) Find its area, correct to the nearest square metre.
(ii) Turf costs $\$12.50$ per square metre. Find the cost of turfing the park, to the nearest dollar.

Solution

Two sides and the angle between them: use the area rule.

$A$	$=$	$\dfrac{1}{2}ab\sin C$
	$=$	$\dfrac{1}{2} \times 46 \times 38 \times \sin 74^\circ 15'$
	$=$	$874\sin 74^\circ 15'$
	$=$	$841.185\ldots$
	$\approx$	$\mathbf{841}\text{ m}^2$

A = \frac{1}{2} \times 46 \times 38 \times \sin 74 ° 15'

(ii) Use the unrounded area:

$\text{cost}$	$=$	$841.185\ldots \times 12.50$
	$\approx$	$\mathbf{\$10\,515}$

Example 2 — Exact areas and working backwards

(i) Find the exact area of a triangle with sides $6$ cm and $8$ cm and an included angle of $135^\circ$.
(ii) Another triangle has sides $7$ cm and $12$ cm and an area of $33.6$ cm$^2$. Find the two possible sizes of the angle $\theta$ between these sides, correct to one decimal place.

Solution

(i) $\sin 135^\circ = \sin 45^\circ = \dfrac{1}{\sqrt{2}}$.

$A$	$=$	$\dfrac{1}{2} \times 6 \times 8 \times \dfrac{1}{\sqrt{2}}$
	$=$	$\dfrac{24}{\sqrt{2}} = \mathbf{12\sqrt{2}}\text{ cm}^2$

(ii) Substitute the area and solve for $\sin\theta$.

$\dfrac{1}{2} \times 7 \times 12 \times \sin\theta$	$=$	$33.6$
$42\sin\theta$	$=$	$33.6$
$\sin\theta$	$=$	$0.8$
$\theta$	$=$	$53.130\ldots^\circ \text{ or } 180^\circ - 53.130\ldots^\circ$
	$\approx$	$\mathbf{53.1^\circ}\text{ or }\mathbf{126.9^\circ}$

42 \sin θ = 33.6

Both angles give a real triangle with the same area, because $\sin(180^\circ - \theta) = \sin\theta$.

Example 3 — Three sides: cosine rule first

Find the area of a triangle with sides $7$ cm, $9$ cm and $12$ cm, correct to one decimal place.

Solution

The area rule needs an angle. Find any angle with the cosine rule, here $\theta$ between the $7$ cm and $9$ cm sides.

$\cos\theta$	$=$	$\dfrac{7^2 + 9^2 - 12^2}{2 \times 7 \times 9} = -\dfrac{14}{126}$
$\theta$	$=$	$96.379\ldots^\circ$
$A$	$=$	$\dfrac{1}{2} \times 7 \times 9 \times \sin 96.379\ldots^\circ$
	$=$	$31.304\ldots$
	$\approx$	$\mathbf{31.3}\text{ cm}^2$

A = \frac{1}{2} \times 7 \times 9 \times \sin θ

Use the sides that form $\theta$ in the area rule. Keep $\theta$ unrounded in the calculator.

Example 4 — Area of a regular polygon

A regular pentagon is drawn with all five vertices on a circle of radius $8$ cm. Find the area of the pentagon, correct to one decimal place.

Solution

Join the centre $O$ to each vertex. This makes $5$ identical triangles, each with two sides of $8$ cm and an angle of $\dfrac{360^\circ}{5} = 72^\circ$ at $O$.

$A_1$	$=$	$\dfrac{1}{2} \times 8 \times 8 \times \sin 72^\circ = 30.433\ldots$
$A$	$=$	$5 \times 30.433\ldots$
	$=$	$152.169\ldots$
	$\approx$	$\mathbf{152.2}\text{ cm}^2$

A = 5 \times \frac{1}{2} \times 8^{2} \times \sin 72 °

In general, a regular $n$-gon in a circle of radius $r$ has area $\dfrac{n}{2}r^2\sin\left(\dfrac{360^\circ}{n}\right)$.

Common pitfalls

Use the included angle. The angle must be between the two sides you multiply. With sides $a$ and $b$, use angle $C$, not $A$ or $B$.

Do not forget the half. $ab\sin C$ is the area of a parallelogram. The triangle is half of it.

Two angles can give the same area. From $\sin\theta = 0.8$, both $53.1^\circ$ and $126.9^\circ$ work. Give both unless the shape rules one out.

Degrees and minutes. $74^\circ 15'$ is $74.25^\circ$, not $74.15^\circ$.

Frequently asked questions

What is the area rule for a triangle?

The area of a triangle is half of two sides times the sine of the angle between them: $A = \dfrac{1}{2}ab\sin C$. It works for any triangle, not just right-angled ones.

How do you find the area of a triangle without the height?

If you know two sides and the angle between them, use A = one half a b sin C. If you know all three sides, use the cosine rule to find an angle first, then the area rule.

How do you find an angle from the area of a triangle?

Substitute the area and the two sides into A = one half a b sin C and solve for sin C. Then find C with inverse sine, and remember that 180 degrees minus C has the same sine, so it may also be an answer.

Does the area rule work for obtuse angles?

Yes. The sine of an obtuse angle is positive and $\sin(180^\circ - C) = \sin C$, so $\sin 130^\circ = \sin 50^\circ$. Substitute the obtuse angle directly.

When is the area of a triangle with two fixed sides the largest?

The area one half a b sin theta is largest when sin theta equals 1, which happens when the angle between the two sides is 90 degrees. The largest area is then one half a b.

\(A\)	\(=\)	\(\dfrac{1}{2}ab\sin C\)
\(\)	\(=\)	\(\dfrac{1}{2} \times 46 \times 38 \times \sin 74^\circ 15'\)
\(\)	\(=\)	\(874\sin 74^\circ 15'\)
\(\)	\(=\)	\(841.185\ldots\)
\(\)	\(\approx\)	\(\mathbf{841}\text{ m}^2\)

\(\text{cost}\)	\(=\)	\(841.185\ldots \times 12.50\)
\(\)	\(\approx\)	\(\mathbf{\$10\,515}\)

\(A\)	\(=\)	\(\dfrac{1}{2} \times 6 \times 8 \times \dfrac{1}{\sqrt{2}}\)
\(\)	\(=\)	\(\dfrac{24}{\sqrt{2}} = \mathbf{12\sqrt{2}}\text{ cm}^2\)

\(\dfrac{1}{2} \times 7 \times 12 \times \sin\theta\)	\(=\)	\(33.6\)
\(42\sin\theta\)	\(=\)	\(33.6\)
\(\sin\theta\)	\(=\)	\(0.8\)
\(\theta\)	\(=\)	\(53.130\ldots^\circ \text{ or } 180^\circ - 53.130\ldots^\circ\)
\(\)	\(\approx\)	\(\mathbf{53.1^\circ}\text{ or }\mathbf{126.9^\circ}\)

\(\cos\theta\)	\(=\)	\(\dfrac{7^2 + 9^2 - 12^2}{2 \times 7 \times 9} = -\dfrac{14}{126}\)
\(\theta\)	\(=\)	\(96.379\ldots^\circ\)
\(A\)	\(=\)	\(\dfrac{1}{2} \times 7 \times 9 \times \sin 96.379\ldots^\circ\)
\(\)	\(=\)	\(31.304\ldots\)
\(\)	\(\approx\)	\(\mathbf{31.3}\text{ cm}^2\)

\(A_1\)	\(=\)	\(\dfrac{1}{2} \times 8 \times 8 \times \sin 72^\circ = 30.433\ldots\)
\(A\)	\(=\)	\(5 \times 30.433\ldots\)
\(\)	\(=\)	\(152.169\ldots\)
\(\)	\(\approx\)	\(\mathbf{152.2}\text{ cm}^2\)

Area Rule

📖 Theory

How to use the area rule

Working backwards and composite shapes

Common pitfalls

Frequently asked questions

Theory