Special Angles and Exact Ratios

Get the exact sin, cos and tan of 30°, 45° and 60° from half a square and half an equilateral triangle, and use them for exact answers with no calculator.

In the exam for Cambridge IGCSE Maths 0580 · 6 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

Why These Angles Are Special

For most angles you need a calculator. For $3 0^{\circ}$ , $4 5^{\circ}$ and $6 0^{\circ}$ you can get sin, cos and tan exactly, as fractions and square roots.

They all come from two triangles you can draw in ten seconds: half a square and half an equilateral triangle.

Key idea

An exact value keeps the square root, like the square root of 3 over 2. A decimal such as 0.866 is only close.

Half a Square: 45 Degrees

Cut a square of side 1 along its diagonal. You get a right-angled triangle with two $4 5^{\circ}$ angles.

Worked example

1
$hyp^{2} = 1^{2} + 1^{2}$ — Both short sides are 1 and the diagonal is the hypotenuse, so use Pythagoras.
2
$hyp^{2} = 2$ — Add the squares: 1 + 1.
3
$hyp = 2$ — Square root. Leave it as a root, because that is the exact value.
4
$sin 4 5^{\circ} = \frac{1}{2}$ — Opposite is 1 and the hypotenuse is $2$ , so sin is opposite over hypotenuse.
5
$\frac{1}{2} \times \frac{2}{2} = \frac{2}{2}$ — Multiply top and bottom by $2$ so there is no root on the bottom. This does not change the value.
6
$cos 4 5^{\circ} = \frac{1}{2} = \frac{2}{2}$ — Adjacent is also 1, so cos is the same as sin.
7
$tan 4 5^{\circ} = \frac{1}{1} = 1$ — Opposite and adjacent are both 1. Tan does not use the hypotenuse, so there is no root.

Common mistake

Writing $tan 4 5^{\circ} = \frac{2}{2}$ . Tan does not use the hypotenuse, so there is no square root: it is 1.

Half an Equilateral Triangle: 30 and 60 Degrees

Take an equilateral triangle of side 2, with all angles $6 0^{\circ}$ . Cut it down the middle. Each half has angles $3 0^{\circ}$ , $6 0^{\circ}$ and $9 0^{\circ}$ .

Worked example

1
$base = 2 \div 2 = 1$ — Cutting the triangle down the middle halves the base of 2.
2
$h^{2} = 2^{2} - 1^{2} = 3$ — The side of 2 is the hypotenuse of each half, so subtract the short side's square from it.
3
$h = 3$ — Square root, kept exact. The sides are now 1, $3$ and 2.
4
$sin 3 0^{\circ} = \frac{1}{2}$ — At the 30° corner the side opposite is the 1 and the hypotenuse is 2.
5
$cos 3 0^{\circ} = \frac{3}{2}$ — At the 30° corner the side next to it is $3$ , and the hypotenuse is 2.
6
$tan 3 0^{\circ} = \frac{1}{3}$ — Opposite is 1 and adjacent is $3$ . There is no hypotenuse in tan.
7
$sin 6 0^{\circ} = \frac{3}{2}$ — At the 60° corner opposite and adjacent swap. Now the side opposite is $3$ .
8
$cos 6 0^{\circ} = \frac{1}{2}$ — The side next to the 60° angle is the 1, with hypotenuse 2.
9
$tan 6 0^{\circ} = \frac{3}{1} = 3$ — Opposite $3$ over adjacent 1.

Common mistake

Swapping the two. The side of 1 is across from the $3 0^{\circ}$ angle, so $sin 3 0^{\circ}$ is the small one, one half.

The Table of Exact Values

Learn this table, or rebuild it from the two triangles.

Angle	sin, cos, tan
$0^{\circ}$	$0, 1, 0$
$3 0^{\circ}$	$\frac{1}{2}, \frac{3}{2}, \frac{1}{3}$
$4 5^{\circ}$	$\frac{2}{2}, \frac{2}{2}, 1$
$6 0^{\circ}$	$\frac{3}{2}, \frac{1}{2}, 3$
$9 0^{\circ}$	1, 0, no value

Memory trick

The sines of 0, 30, 45, 60 and 90 degrees are $\frac{0}{2}, \frac{1}{2}, \frac{2}{2}, \frac{3}{2}, \frac{4}{2}$ . The cosines are the same list backwards.

Exact Answers Without a Calculator

The exact-value question: find a side and leave the square root in. The hypotenuse is 10 cm and the angle is $6 0^{\circ}$ . Find the opposite side exactly.

Worked example

1
$sin 6 0^{\circ} = \frac{opp}{10}$ — The side wanted is opposite the 60° angle and 10 is the hypotenuse, so use sin.
2
$opp = 10 \times sin 6 0^{\circ}$ — The unknown is on top, so multiply both sides by 10.
3
$opp = 10 \times \frac{3}{2}$ — The question says exact, so swap sin 60° for its exact value from the table, not a decimal.
4
$opp = \frac{10 3}{2}$ — Multiply 10 by the top of the fraction.
5
$opp = 53 cm$ — Divide 10 by 2 and leave the root in. That is the exact answer.

It works backwards too. If $sin x = \frac{1}{2}$ , read the table: $x = 3 0^{\circ}$ .

Common mistake

Writing 8.66 cm when the question says exact. Leave it as $53$ .

Dividing by an Exact Value

Sometimes the side you want is on the bottom of the fraction. Then you divide by an exact value and a root lands on the bottom. Exact answers should have no root on the bottom, so multiply the top and the bottom by that root.

Formula

\frac{a}{b} = \frac{a b}{b}

A right-angled triangle has a $6 0^{\circ}$ angle. The side opposite it is 12 cm. Find the side next to the angle, exactly, in the form $a b$ .

Worked example

1
$tan 6 0^{\circ} = \frac{12}{x}$ — 12 is opposite the 60° angle and $x$ is next to it, so use tan.
2
$3 = \frac{12}{x}$ — Swap tan 60° for its exact value from the table.
3
$x = \frac{12}{3}$ — $x$ is on the bottom, so swap places: divide 12 by $3$ .
4
$x = \frac{12}{3} \times \frac{3}{3}$ — A root should not stay on the bottom. Multiply top and bottom by $3$ , which is the same as multiplying by 1.
5
$x = \frac{12 3}{3}$ — On the bottom, $3 \times 3 = 3$ . On the top, 12 keeps its root.
6
$x = 43 cm$ — Divide 12 by 3. This is the form $a b$ the question asks for.

Common mistake

Turning $\frac{12}{3}$ into 4 by dividing 12 by 3. The root does not disappear: the answer is $43$ , about 6.93 cm.

Key idea

$\frac{1}{3}$ and $\frac{3}{3}$ are the same number, about 0.577. So $tan 3 0^{\circ}$ can be written either way.

Exact Values in Calculations

Some questions are a calculation with no triangle. Swap each ratio for its exact value from the table, then simplify. No calculator.

Worked example

1
$y = sin 6 0^{\circ} \times tan 3 0^{\circ}$ — Find $y$ exactly.
2
$y = \frac{3}{2} \times \frac{1}{3}$ — Swap sin 60° and tan 30° for their exact values from the table.
3
$y = \frac{3 \times 1}{2 \times 3}$ — Multiply the tops together and the bottoms together.
4
$y = \frac{1}{2}$ — $3$ is on the top and the bottom, so it cancels.

Now one with a sum. Find the exact value of $\frac{6 cos 3 0 ^{\circ}}{3} + tan 4 5^{\circ}$ .

Worked example

1
$\frac{6 \times \frac{3}{2}}{3} + 1$ — Swap cos 30° and tan 45° for their exact values.
2
$\frac{3 3}{3} + 1$ — 6 times $\frac{3}{2}$ is $33$ , because $6 \div 2 = 3$ .
3
$3 + 1 = 4$ — $3$ cancels on the top and the bottom, leaving 3. Then add 1.

Common mistake

Squaring only the top. $(\frac{3}{2})^{2} = \frac{3}{4}$ , because the top and the bottom are both squared.

Which Method? Read the Question

Check the angle and whether the question says exact.

You see	You do
30, 45 or 60 degrees and the word exact	use the table, keep the square root
A ratio like one half, find the angle	read the table backwards
Any other angle	Choosing the Right Trigonometric Ratio
Only two sides, no angle	Pythagoras' Theorem
Angles past 90 degrees	Trigonometric Graphs and Equations
A calculation such as sin 60° × tan 30°	swap each ratio for its exact value, then simplify
An exact length with the unknown on the bottom	divide, then remove the root from the bottom

Key idea

If you forget a value, sketch the triangle and use Pythagoras for the missing side.

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Real past-paper questions on special angles and exact ratios

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