Exact values from special triangles | Year 10

01 / LEARN

Understand the idea.

Build the exact ratios from triangles

Exact trigonometric values come from special triangles. Splitting a unit square diagonally gives side ratio $1 : 1 : \sqrt{2}$ ; halving an equilateral triangle of side $2$ gives $1 : \sqrt{3} : 2$ .

The shortest side in the second triangle faces $30^{\circ}$ and the $\sqrt{3}$ side faces $60^{\circ}$ . Label the angle, then use opposite, adjacent and hypotenuse to build the required ratio.

Choose the ratio before substituting

The function matters as much as the angle: $sin 60^{\circ} = \frac{\sqrt{3}}{2}$ whereas $tan 60^{\circ} = \sqrt{3}$ . The denominator changes because sine uses the hypotenuse and tangent uses the adjacent leg.

Scaling every side by the same factor preserves these ratios. Keep square roots exact when requested; a rounded decimal is a check rather than an equivalent exact answer. Simplify surds after multiplying.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Exact ratios

Find cos 30° and then simplify tan 60° × sin 30°.

Worked solution

1
Read cosine from the special triangle.
$cos 30^{\circ} = \frac{\sqrt{3}}{2}$
Cosine is adjacent divided by hypotenuse in the $1 : \sqrt{3} : 2$ triangle.
2
Write the exact factors in the product.
$tan 60^{\circ} = \sqrt{3}, sin 30^{\circ} = \frac{1}{2}$
Tangent and sine use different side ratios.
3
Multiply the exact values.
$tan 60^{\circ} sin 30^{\circ} = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$
No decimal approximation is needed.

Answer

$cos 30^{\circ} = \sqrt{3} / 2$ ; product $\sqrt{3} / 2$

EXAMPLE 02

Scale a special triangle

A right triangle has hypotenuse 12 cm and an angle of 60°. Find both legs exactly.

Worked solution

1
Use cosine for the adjacent leg.
$a = 12 cos 60^{\circ} = 12 \cdot \frac{1}{2} = 6 cm$
The hypotenuse is known and the adjacent leg is required.
2
Use sine for the opposite leg.
$b = 12 sin 60^{\circ} = 12 \cdot \frac{\sqrt{3}}{2} = 6 \sqrt{3} cm$
Sine compares the opposite leg with the hypotenuse.
3
Check the exact lengths.
$6^{2} + {(6 \sqrt{3})}^{2} = 36 + 108 = 144 = 12^{2}$
The two legs satisfy Pythagoras with the stated hypotenuse.

Answer

$6$ cm and $6 \sqrt{3}$ cm

03 / YOUR TURN

Try it yourself.

Check your understanding

Find sin 45° exactly.

Show answer & reasoning

$\sqrt{2} / 2$

The isosceles right triangle gives $sin 45^{\circ} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ . Multiplying numerator and denominator by $\sqrt{2}$ preserves the value.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Find $sin 30^{\circ}$ exactly.
2 marks
Find $cos 45^{\circ}$ exactly.
2 marks
Find $tan 30^{\circ}$ exactly.
2 marks
A right triangle has hypotenuse 10 cm and an angle of $45^{\circ}$ . Find the opposite leg exactly.
3 marks
Evaluate $2 cos 60^{\circ} + sin 30^{\circ}$ exactly.
2 marks
Are $sin 60^{\circ}$ and $tan 60^{\circ}$ equal? Give exact values to justify.
2 marks

Read the worked solutions

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

Find $sin 30^{\circ}$ exactly.
2 marks
The shorter leg is half the hypotenuse.
$\frac{1}{2}$
Find $cos 45^{\circ}$ exactly.
2 marks
$cos 45^{\circ} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ .
$\frac{\sqrt{2}}{2}$
Find $tan 30^{\circ}$ exactly.
2 marks
$tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ .
$\frac{\sqrt{3}}{3}$
A right triangle has hypotenuse 10 cm and an angle of $45^{\circ}$ . Find the opposite leg exactly.
3 marks
$x = 10 sin 45^{\circ} = 10 (\frac{\sqrt{2}}{2}) = 5 \sqrt{2}$ cm.
$5 \sqrt{2}$ cm
Evaluate $2 cos 60^{\circ} + sin 30^{\circ}$ exactly.
2 marks
$2 (\frac{1}{2}) + \frac{1}{2} = \frac{3}{2}$ .
$\frac{3}{2}$
Are $sin 60^{\circ}$ and $tan 60^{\circ}$ equal? Give exact values to justify.
2 marks
$sin 60^{\circ} = \frac{\sqrt{3}}{2}$ whereas $tan 60^{\circ} = \sqrt{3}$ .
The ratios have different denominators.
No.

Checking saved progress…

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.

Build the exact ratios from triangles

Choose the ratio before substituting

See the method in action.

Exact ratios

Read cosine from the special triangle.

Write the exact factors in the product.

Multiply the exact values.

Scale a special triangle

Use cosine for the adjacent leg.

Use sine for the opposite leg.

Check the exact lengths.

Try it yourself.

Check your understanding

Keep practising