Area of non-right triangles | Year 10

01 / LEARN

Understand the idea.

Two sides and the angle between them

The sine area formula is half base times perpendicular height written using an angle. If sides $a$ and $b$ meet at $C$ , the height relative to base $a$ is $b sin C$ .

The selected angle must lie between the two selected sides. A different angle does not describe that height. The formula also works for an obtuse included angle because its sine equals the sine of its acute supplement.

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

Use the same relationship backwards

The same relationship can find a side or angle from a known area. Isolate the unknown side, or isolate the sine of the angle, before using a calculator. Length answers and area answers require different units.

When finding an included angle, check both inverse-sine candidates. An acute angle and its obtuse supplement give equal heights and areas for the same two sides, unless another condition rules one out.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Find an area

Two sides of a triangle are 6 m and 9 m. Their included angle is 40°. Find the area to 0.1 square metre.

Worked solution

1
Select the included-angle area formula.
$A = \frac{1}{2} (6) (9) sin 40^{\circ}$
The $40^{\circ}$ angle lies between the two known sides.
2
Evaluate without rounding the sine.
$A = 27 sin 40^{\circ} \approx 17.3553 m^{2}$
The sine factor supplies the perpendicular-height adjustment.
3
Round the area to the requested precision.
$A \approx 17.4 m^{2}$
Area is reported to one decimal place in square metres.

Answer

17.4 m²

EXAMPLE 02

Two angles with the same area

A triangle has area 24 cm² and two sides 8 cm and 10 cm. Find all possible included angles to 0.1°.

Worked solution

1
Substitute the known area and sides.
$24 = \frac{1}{2} (8) (10) sin C = 40 sin C$
$C$ is the angle included between the $8 cm$ and $10 cm$ sides.
2
Isolate its sine.
$sin C = \frac{24}{40} = 0.6$
Divide by the product multiplying the sine.
3
Find the acute angle and supplement.
$\begin{aligned} C_{1} & = {sin}^{- 1} (0.6) \approx {36.9}^{\circ} \\ C_{2} & = 180^{\circ} - C_{1} \approx {143.1}^{\circ} \end{aligned}$
Both angles lie inside a triangle and have sine $0.6$ ; neither is excluded by the question.

Answer

36.9° or 143.1°

03 / YOUR TURN

Try it yourself.

Check your understanding

Sides 6 cm and 8 cm include 90°. Find the area.

Show answer & reasoning

24 cm²

$A = \frac{1}{2} (6) (8) sin 90^{\circ} = 24 {cm}^{2}$ . With a right included angle, the second side already is the perpendicular height.

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

Find the area when sides 10 cm and 12 cm include $30^{\circ}$ .
2 marks
Sides 14 m and 9 m include $50^{\circ}$ . Find the area to 0.1 m².
2 marks
A triangle has area 15 cm². One side is 6 cm and another side $b$ meets it at $30^{\circ}$ . Find $b$ .
3 marks
Sides 5 cm and 10 cm enclose area 12.5 cm². Find both possible included angles.
3 marks
Two sides of a triangular sign are 8 cm and 11 cm. Their included angle is acute and the area is 22 cm². Find that angle.
3 marks
$A B = 7$ cm, $A C = 9$ cm and angle $B = 40^{\circ}$ . Explain why $\frac{1}{2} \times 7 \times 9 \times sin 40^{\circ}$ is not justified.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the area when sides 10 cm and 12 cm include $30^{\circ}$ .
2 marks
$A = \frac{1}{2} (10) (12) sin 30^{\circ} = 30$ cm².
30 cm²
Sides 14 m and 9 m include $50^{\circ}$ . Find the area to 0.1 m².
2 marks
$A = \frac{1}{2} (14) (9) sin 50^{\circ}$ .
48.3 m²
A triangle has area 15 cm². One side is 6 cm and another side $b$ meets it at $30^{\circ}$ . Find $b$ .
3 marks
$15 = \frac{1}{2} (6) b sin 30^{\circ} = 1.5 b$ .
$b = 10$ cm.
10 cm
Sides 5 cm and 10 cm enclose area 12.5 cm². Find both possible included angles.
3 marks
$sin C = \frac{12.5}{25} = \frac{1}{2}$ .
The two angles between $0^{\circ}$ and $180^{\circ}$ are $30^{\circ}$ and $150^{\circ}$ .
$30^{\circ}$ or $150^{\circ}$
Two sides of a triangular sign are 8 cm and 11 cm. Their included angle is acute and the area is 22 cm². Find that angle.
3 marks
$sin C = \frac{22}{44} = \frac{1}{2}$ .
The acute answer is $30^{\circ}$ .
$30^{\circ}$
$A B = 7$ cm, $A C = 9$ cm and angle $B = 40^{\circ}$ . Explain why $\frac{1}{2} \times 7 \times 9 \times sin 40^{\circ}$ is not justified.
2 marks
The selected sides meet at $A$ , so their included angle is $A$ .
Find $A$ before using those two sides in the area rule.
$40^{\circ}$ is not the angle between $A B$ and $A C$ .

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Curriculum reference & lesson scope

Australian Curriculum v9 and South Australian school course scope

Right-triangle applications include AC9M10M03. The approved chapter also develops non-right-triangle methods, exact values and trigonometric functions within one normal Year 10 course.

Official curriculum source ↗

In this lesson

Understand the idea.

Two sides and the angle between them

Use the same relationship backwards

See the method in action.

Find an area

Select the included-angle area formula.

Evaluate without rounding the sine.

Round the area to the requested precision.

Two angles with the same area

Substitute the known area and sides.

Isolate its sine.

Find the acute angle and supplement.

Try it yourself.

Check your understanding

Keep practising