Non-right-triangle area | Year 11

01 / LEARN

Understand the idea.

From base and height

Take one known side as a base and resolve the other side into a perpendicular height. That height is $b sin C$ , so half base times height gives the area formula. It also works for obtuse included angles using an extended base.

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

Bounds and inverse questions

For fixed side lengths, the greatest height occurs when the sides meet at a right angle. Thus the area cannot exceed $\frac{1}{2} a b$ . An area below this maximum usually corresponds to two supplementary included angles unless the wording selects one.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Sides 8 cm and 12 cm enclose 120°. Find the exact area.

Worked solution

1
Use the two sides and their included angle.
$K = \frac{1}{2} (8) (12) sin 120^{\circ} = 48 sin 120^{\circ}$
The given angle is between the two sides in the area product.
2
Find the exact sine.
$sin 120^{\circ} = sin 60^{\circ} = \frac{\sqrt{3}}{2}$
An obtuse angle and its acute supplement have the same sine.
3
Calculate the area with square units.
$K = 48 (\frac{\sqrt{3}}{2}) = 24 \sqrt{3} {cm}^{2}$
Multiplying two lengths produces an area, so the unit is square centimetres.

Answer

$24 \sqrt{3} {cm}^{2}$

EXAMPLE 02

Worked example 2

Sides 10 m and 12 m enclose angle C. The area is 30 m². Find possible C.

Worked solution

1
Substitute the known area.
$30 = \frac{1}{2} (10) (12) sin C = 60 sin C$
The area formula uses the unknown angle between the given sides.
2
Isolate its sine.
$sin C = \frac{30}{60} = \frac{1}{2}$
The ratio is within the possible sine range for a triangle angle.
3
Keep both possible included angles.
$C = 30^{\circ} or C = 150^{\circ}$
Each angle has sine one half, and no acute or obtuse condition selects just one.

Answer

30° or 150°

03 / YOUR TURN

Try it yourself.

Check your understanding

What is the greatest area between sides 5 cm and 9 cm?

Show answer & reasoning

22.5 cm².

The largest possible sine is 1, attained at $90^{\circ}$ . Therefore $K_{max} = \frac{1}{2} (5) (9) = 22.5 {cm}^{2}$ .

Keep practising

6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

Find area from sides 7 cm,10 cm and included angle $30^{\circ}$ .
3 marks
Find area from sides 6 cm,8 cm and included angle $60^{\circ}$ exactly.
3 marks
Find exact area for sides 9 m,14 m and included angle $45^{\circ}$ .
3 marks
Sides 4 cm and 10 cm enclose an obtuse angle. Area is 10 cm². Find the angle.
3 marks
A sail has sides 3.5 m and 4 m meeting at $75^{\circ}$ . Find area to two decimal places.
3 marks
Can sides 6 cm and 8 cm enclose area 25 cm²? Explain.
3 marks

Read the worked solutions

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

Find area from sides 7 cm,10 cm and included angle $30^{\circ}$ .
3 marks
$K = 35 (\frac{1}{2}) = 17.5$ .
17.5 cm²
Find area from sides 6 cm,8 cm and included angle $60^{\circ}$ exactly.
3 marks
$K = 24 (\frac{\sqrt{3}}{2})$ .
$12 \sqrt{3} {cm}^{2}$
Find exact area for sides 9 m,14 m and included angle $45^{\circ}$ .
3 marks
$K = 63 sin 45^{\circ} = \frac{63 \sqrt{2}}{2}$ .
$\frac{63 \sqrt{2}}{2} m^{2}$
Sides 4 cm and 10 cm enclose an obtuse angle. Area is 10 cm². Find the angle.
3 marks
$sin C = \frac{10}{20} = \frac{1}{2}$ .
Select the obtuse supplementary solution.
$150^{\circ}$ .
A sail has sides 3.5 m and 4 m meeting at $75^{\circ}$ . Find area to two decimal places.
3 marks
$K = 7 sin 75^{\circ} \approx 6.76148$ .
6.76 m²
Can sides 6 cm and 8 cm enclose area 25 cm²? Explain.
3 marks
$K \leq \frac{1}{2} (6) (8) = 24$ .
No; maximum area is 24 cm².

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: triangle rules, circular measure and trigonometric functions; school sequencing may vary.

Official curriculum source ↗

In this lesson

Understand the idea.

From base and height

Bounds and inverse questions

See the method in action.

Worked example 1

Use the two sides and their included angle.

Find the exact sine.

Calculate the area with square units.

Worked example 2

Substitute the known area.

Isolate its sine.

Keep both possible included angles.

Try it yourself.

Check your understanding

Keep practising