These values must be known by heart. They come up on non-calculator papers and in questions that ask for an exact (surd) answer.
| $\theta$ | $0^\circ$ | $30^\circ$ | $45^\circ$ | $60^\circ$ | $90^\circ$ |
|---|---|---|---|---|---|
| $\sin\theta$ | $0$ | $\dfrac{1}{2}$ | $\dfrac{\sqrt{2}}{2}$ | $\dfrac{\sqrt{3}}{2}$ | $1$ |
| $\cos\theta$ | $1$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{\sqrt{2}}{2}$ | $\dfrac{1}{2}$ | $0$ |
| $\tan\theta$ | $0$ | $\dfrac{1}{\sqrt{3}}$ | $1$ | $\sqrt{3}$ | undefined |
You do not have to memorise the table blindly — every value can be derived from two special triangles. Learning to draw them takes a minute and makes the whole table recoverable.
Triangle 1 — the $45^\circ$ triangle
Take a square of side $1$ and cut it along a diagonal.
$\sin 45^\circ = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$, $\cos 45^\circ = \dfrac{1}{\sqrt{2}}$, $\tan 45^\circ = \dfrac{1}{1} = 1$.
Triangle 2 — the $30^\circ$–$60^\circ$ triangle
Take an equilateral triangle of side $2$ and cut it in half.
$\sin 30^\circ = \dfrac{1}{2}$, $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$, $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$
$\sin 60^\circ = \dfrac{\sqrt{3}}{2}$, $\cos 60^\circ = \dfrac{1}{2}$, $\tan 60^\circ = \dfrac{\sqrt{3}}{1} = \sqrt{3}$
• $\sin$ goes $0$, $\tfrac{1}{2}$, $\tfrac{\sqrt2}{2}$, $\tfrac{\sqrt3}{2}$, $1$ — you can write it as $\dfrac{\sqrt{0}}{2}, \dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{4}}{2}$.
• $\cos$ is the same list backwards.
• $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$, so you can always work tan out from the other two.
A right-angled triangle has hypotenuse $12$ cm and an angle of $60^\circ$. Find the exact length of the opposite side.
(As a decimal that is $10.39$ cm, but $6\sqrt{3}$ is the exact answer.)
An equilateral triangle has side $8$ cm. Find its exact area.
Check: $16\sqrt{3} = 27.7\text{ cm}^2$ ✓
A right-angled triangle has an angle of $30^\circ$ and an opposite side of $5$ cm. Find the exact adjacent side.
$A = 5\sqrt{3}$ cm — which is $8.66$ cm.
In a right-angled triangle the opposite side is $7$ cm and the hypotenuse is $14$ cm. Find the angle without a calculator.
In a right-angled triangle, $\tan\theta = \sqrt{3}$. Find $\theta$.
| Expression | Simplified |
|---|---|
| $\dfrac{1}{\sqrt{2}}$ | $\dfrac{\sqrt{2}}{2}$ (multiply top and bottom by $\sqrt{2}$) |
| $\dfrac{1}{\sqrt{3}}$ | $\dfrac{\sqrt{3}}{3}$ |
| $\sqrt{3} \times \sqrt{3}$ | $3$ |
| $\left(\dfrac{\sqrt{3}}{2}\right)^2$ | $\dfrac{3}{4}$ |
| $10 \times \dfrac{\sqrt{2}}{2}$ | $5\sqrt{2}$ |
Show that $\sin^2 60^\circ + \cos^2 60^\circ = 1$.
This works for every angle — it is a direct consequence of Pythagoras.
$30^\circ$
$\sin = \tfrac{1}{2}$, $\cos = \tfrac{\sqrt3}{2}$, $\tan = \tfrac{1}{\sqrt3}$.
$45^\circ$
$\sin = \cos = \tfrac{\sqrt2}{2}$, $\tan = 1$.
$60^\circ$
$\sin = \tfrac{\sqrt3}{2}$, $\cos = \tfrac{1}{2}$, $\tan = \sqrt3$.
$0^\circ$ and $90^\circ$
$\sin 0 = 0$, $\cos 0 = 1$; $\sin 90 = 1$, $\cos 90 = 0$.
$\tan 90^\circ$
Undefined — the calculator errors.
The two triangles
Half a unit square, and half an equilateral triangle of side $2$.
The pattern
$\sin$: $\tfrac{\sqrt0}{2}, \tfrac{\sqrt1}{2}, \tfrac{\sqrt2}{2}, \tfrac{\sqrt3}{2}, \tfrac{\sqrt4}{2}$. $\cos$ is the reverse.
Link
$\tan\theta = \dfrac{\sin\theta}{\cos\theta}$.
Exact means surds
Leave $\sqrt{3}$ in the answer; do not turn it into $1.732$.
Write down the exact values of (a) $\sin 30^\circ$, (b) $\cos 60^\circ$, (c) $\tan 45^\circ$.
▶ Show solution
(a) $\dfrac{1}{2}$ (b) $\dfrac{1}{2}$ (c) $1$
Write down the exact values of (a) $\cos 30^\circ$, (b) $\sin 60^\circ$, (c) $\tan 60^\circ$.
▶ Show solution
(a) $\dfrac{\sqrt{3}}{2}$ (b) $\dfrac{\sqrt{3}}{2}$ (c) $\sqrt{3}$
Without a calculator, find $\theta$ if (a) $\cos\theta = \dfrac{1}{2}$, (b) $\tan\theta = 1$, (c) $\sin\theta = 1$.
▶ Show solution
(a) $\theta = 60^\circ$ (b) $\theta = 45^\circ$ (c) $\theta = 90^\circ$
A right-angled triangle has hypotenuse $20$ cm and an angle of $30^\circ$. Find the exact length of the side opposite that angle.
▶ Show solution
$\sin 30^\circ = \dfrac{O}{20}$
$O = 20 \times \dfrac{1}{2} = 10$ cm
A right-angled triangle has an angle of $45^\circ$ and an adjacent side of $6$ cm. Find the exact hypotenuse.
▶ Show solution
$\cos 45^\circ = \dfrac{6}{H}$
$H = \dfrac{6}{\cos 45^\circ} = \dfrac{6}{\frac{\sqrt{2}}{2}} = \dfrac{12}{\sqrt{2}}$
Rationalise: $\dfrac{12}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{12\sqrt{2}}{2} = 6\sqrt{2}$ cm
Find the exact height of an equilateral triangle of side $10$ cm.
▶ Show solution
Halving the base gives a right-angled triangle with hypotenuse $10$ and base $5$.
$h = 10 \sin 60^\circ = 10 \times \dfrac{\sqrt{3}}{2} = 5\sqrt{3}$ cm
Check with Pythagoras: $\sqrt{100 - 25} = \sqrt{75} = 5\sqrt{3}$ ✓
A right-angled triangle has an angle of $60^\circ$ with an adjacent side of $9$ cm. Find the exact length of the opposite side.
▶ Show solution
$\tan 60^\circ = \dfrac{O}{9}$
$O = 9 \times \sqrt{3} = 9\sqrt{3}$ cm
Evaluate exactly: $\;2\sin 30^\circ + 3\cos 60^\circ - \tan 45^\circ$.
▶ Show solution
$2 \times \dfrac{1}{2} + 3 \times \dfrac{1}{2} - 1$
$= 1 + 1.5 - 1 = 1.5$ (or $\dfrac{3}{2}$)
An isosceles triangle has two sides of $6$ cm with a $120^\circ$ angle between them. Using the formula Area $= \tfrac{1}{2}ab\sin C$ and the fact that $\sin 120^\circ = \sin 60^\circ$, find the exact area.
▶ Show solution
Area $= \tfrac{1}{2} \times 6 \times 6 \times \sin 120^\circ$
$\sin 120^\circ = \sin 60^\circ = \dfrac{\sqrt{3}}{2}$
Area $= 18 \times \dfrac{\sqrt{3}}{2} = 9\sqrt{3}\text{ cm}^2$
(That is about $15.6\text{ cm}^2$.)
A square-based pyramid has base edges $6$ cm. Each sloping face makes an angle of $60^\circ$ with the base.
(a) Find the exact perpendicular height of the pyramid. (b) Find the exact volume. (c) Find the exact slant height of a triangular face.
▶ Show solution
(a) The angle between a face and the base is measured in the vertical triangle running from the centre of the base to the midpoint of an edge.
That horizontal distance is half the base edge: $6 \div 2 = 3$ cm.
$\tan 60^\circ = \dfrac{h}{3}$, so $h = 3\sqrt{3}$ cm.
(b) Base area $= 6 \times 6 = 36\text{ cm}^2$.
$V = \tfrac{1}{3} \times 36 \times 3\sqrt{3} = 36\sqrt{3}\text{ cm}^3$
(About $62.4\text{ cm}^3$.)
(c) The slant height $l$ is the hypotenuse of that same triangle:
$\cos 60^\circ = \dfrac{3}{l}$, so $l = \dfrac{3}{\frac{1}{2}} = 6$ cm.