NSWMaths Standard 2Syllabus dot point

How is the area of a non-right-angled triangle calculated using two sides and the included angle?

Use the formula $A = \frac{1}{2} a b \sin C$ to find the area of any triangle given two sides and the included angle

A focused answer to the HSC Maths Standard 2 dot point on the area formula $A = \frac{1}{2} a b \sin C$ . When to use it, how it derives from the standard $\frac{1}{2}$ base times height formula, and worked Australian land surveying examples.

Generated by Claude OpusReviewed by Better Tuition Academy6 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to use the formula $A = \frac{1}{2} a b \sin C$ to find the area of a triangle whenever you have two sides and the angle between them. This is the only triangle area formula you need beyond the standard $\frac{1}{2} \times \text{base} \times \text{height}$ .

The answer

The formula

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

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

The two sides must be the ones forming the chosen angle (the included angle).

Why this works

Drop a perpendicular from vertex $A$ to side $BC$ . The height of the triangle is $h = b \sin C$ (right-triangle trigonometry inside the triangle).

Then $A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} a \times b \sin C$ .

The formula is just the base-times-height formula with the height expressed in terms of the side and the angle.

When to use it

You have two sides and the included angle: direct application.
You have three sides (SSS): use the cosine rule first to find an angle, then apply this formula. Or use Heron's formula if you remember it (Heron's is not on the Standard 2 reference sheet, so the cosine-rule path is expected).
You have two angles and one side (AAS): use the sine rule to find the second side, then apply this formula.
You have a right-angled triangle: use $A = \frac{1}{2} \times \text{base} \times \text{height}$ directly.

Units

If sides are in metres, area is in m $^2$ . If in centimetres, area is in cm $^2$ . Always include units in your final answer.

Worked example

Direct application

Triangle with sides $10$ cm, $12$ cm and included angle $40\degree$ .

$A = \frac{1}{2} \times 10 \times 12 \times \sin 40\degree = 60 \times 0.6428 \approx 38.57$ cm $^2$ .

Australian property boundary

A surveyor measures a triangular block of land in regional NSW with sides $42$ m and $38$ m meeting at an angle of $85\degree$ .

$A = \frac{1}{2} \times 42 \times 38 \times \sin 85\degree = 798 \times 0.9962 \approx 794.95$ m $^2$ .

For a quick conversion to hectares: $794.95$ m $^2$ $\approx 0.0795$ ha.

Two-step from SSS

Triangle with sides $9$ m, $12$ m and $15$ m. Find the area.

First check: $9^2 + 12^2 = 81 + 144 = 225 = 15^2$ . It is right-angled with the right angle between the $9$ m and $12$ m sides. So $\sin C = \sin 90\degree = 1$ .

$A = \frac{1}{2} \times 9 \times 12 \times 1 = 54$ m $^2$ .

(Alternatively, recognise the $3$ - $4$ - $5$ triple scaled by $3$ .)

For a non-right SSS triangle, use the cosine rule to find one angle first:

Sides $7$ , $8$ , $9$ . Find area.

$\cos A = \frac{8^2 + 9^2 - 7^2}{2 \times 8 \times 9} = \frac{64 + 81 - 49}{144} = \frac{96}{144} \approx 0.6667$ , so $A \approx 48.19\degree$ .

$A_{\text{area}} = \frac{1}{2} \times 8 \times 9 \times \sin 48.19\degree = 36 \times 0.7454 \approx 26.83$ m $^2$ .

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q152 marksFind the area of a triangle with two sides of length

14

cm and

18

cm meeting at an angle of

50\degree

Show worked answer →

Use $A = \frac{1}{2} a b \sin C$ with $a = 14$ , $b = 18$ , $C = 50\degree$ .

$A = \frac{1}{2} \times 14 \times 18 \times \sin 50\degree$ .

$\sin 50\degree \approx 0.7660$ .

$A = 126 \times 0.7660 \approx 96.52$ cm $^2$ .

Markers reward the formula stated, correct substitution, and an answer rounded to two decimal places or sensibly.

2023 HSC Q203 marksA triangular sail has sides

4.5

m and

3.8

m and an angle of

75\degree

between them. Find the area of the sail correct to one decimal place.

Show worked answer →

$A = \frac{1}{2} \times 4.5 \times 3.8 \times \sin 75\degree$ .

$\sin 75\degree \approx 0.9659$ .

$A = 0.5 \times 4.5 \times 3.8 \times 0.9659 = 8.55 \times 0.9659 \approx 8.26$ m $^2$ .

Rounded to one decimal place: $A \approx 8.3$ m $^2$ .

Markers reward the formula, substitution of the included angle (not just any angle), and the final answer at the requested precision.

What this dot point is asking

The answer

The formula

Why this works

When to use it

Units

Past exam questions, worked

Related dot points