Pythagoras and right-angled trigonometry

Pythagoras’ theorem connects the three side lengths of a right-angled triangle. The trigonometric ratios connect its side lengths to its acute angles. Together, they let you find missing lengths and angles, justify geometric results and model practical situations.

The central skill is not memorising formulas. It is identifying the hypotenuse and labelling the other two sides relative to the angle being used.

Prerequisites

You should be able to:

• square positive numbers and evaluate square roots;
• rearrange simple equations;
• work with fractions and surds;
• recognise a right angle and use the angle sum of a triangle;
• use a scientific calculator in degree mode.

Review rearranging formulae, indices, roots and surds or calculator fluency if these techniques interrupt your geometry.

Reading a right-angled triangle

The hypotenuse is opposite the right angle. It is always the longest side. Relative to a chosen acute angle $\theta$:

• the opposite side is across from $\theta$;
• the adjacent side touches $\theta$ but is not the hypotenuse;
• the hypotenuse does not change when the chosen acute angle changes.

The labels opposite and adjacent are therefore not permanent names. If you switch from one acute angle to the other, those two labels switch.

Label before calculating

Mark the right angle, circle the angle you are using, then write $H$, $O$ and $A$ beside the relevant sides. This prevents most trigonometry errors.

Pythagoras’ theorem

For a right-angled triangle with shorter sides $a$ and $b$ and hypotenuse $c$,

$$\boxed{a^2+b^2=c^2}.$$

The theorem applies only to right-angled triangles. The isolated term $c^2$ must represent the hypotenuse.

Why the theorem is plausible

Construct a square on each side of the triangle. Pythagoras’ theorem says that the combined area of the two smaller squares equals the area of the square on the hypotenuse:

$a^2+b^2=c^2.$

This area interpretation explains both the squares and why the hypotenuse has a special role.

Worked example: find the hypotenuse

A right-angled triangle has shorter sides $7$ cm and $11$ cm. Find its hypotenuse $c$.

$$\begin{aligned} c^2&=7^2+11^2\\ &=49+121\\ &=170,\\ c&=\sqrt{170}\text{ cm}. \end{aligned}$$

This exact answer is approximately $13.0$ cm to three significant figures. The positive square root is used because a length is positive.

Worked example: find a shorter side

The hypotenuse is $13$ cm and one shorter side is $5$ cm. Find the remaining side $x$.

$$\begin{aligned} x^2+5^2&=13^2\\ x^2&=169-25\\ &=144,\\ x&=12\text{ cm}. \end{aligned}$$

When finding a shorter side, subtract the known shorter side squared from the hypotenuse squared. The calculation $5^2-13^2$ would produce a negative value and signals that the hypotenuse has been misidentified.

Testing whether a triangle is right-angled

Put the longest side in the role of $c$. A triangle with side lengths $a$, $b$ and $c$, where $c$ is longest, is right-angled precisely when

$a^2+b^2=c^2.$

For sides $8$, $15$ and $17$,

$8^2+15^2=64+225=289=17^2,$

so the triangle is right-angled.

For sides $6$, $7$ and $9$,

$6^2+7^2=85\ne81=9^2,$

so it is not right-angled.

Sine, cosine and tangent

For an acute angle $\theta$ in a right-angled triangle,

$$\boxed{\sin\theta=\frac{O}{H}},\qquad \boxed{\cos\theta=\frac{A}{H}},\qquad \boxed{\tan\theta=\frac{O}{A}}.$$

The mnemonic SOH CAH TOA records the three ratios. A more durable way to remember them is to notice that sine and cosine involve the hypotenuse, while tangent compares the two shorter sides.

These ratios depend only on the angle, not on the size of the triangle. All right-angled triangles containing the same acute angle are similar, so corresponding side lengths scale by the same factor and their ratios remain unchanged.

Choosing a ratio

After labelling the sides relative to $\theta$:

1. identify the known side and the side to be found;
2. choose the ratio containing exactly those two labels;
3. substitute values before rearranging;
4. check that the result is geometrically reasonable.

For example, if opposite and hypotenuse are involved, use sine. There is no need to write all three ratios and guess.

Worked example: find a side using sine

In a right-angled triangle, an angle is $38^\circ$ and the hypotenuse is $14$ cm. Find the side $x$ opposite the angle.

Opposite and hypotenuse are involved, so use sine:

$$\begin{aligned} \sin38^\circ&=\frac{x}{14}\\ x&=14\sin38^\circ\\ &=8.62\text{ cm}\quad\text{to 3 s.f.} \end{aligned}$$

The answer is shorter than $14$ cm, as every non-hypotenuse side must be.

Worked example: rearrange cosine carefully

An angle is $52^\circ$ and its adjacent side is $9$ m. Find the hypotenuse $h$.

$$\begin{aligned} \cos52^\circ&=\frac9h\\ h\cos52^\circ&=9\\ h&=\frac9{\cos52^\circ}\\ &=14.6\text{ m}\quad\text{to 3 s.f.} \end{aligned}$$

Multiplying by $\cos52^\circ$ would give a result below $9$ m, impossible for the hypotenuse. This size check catches the common rearrangement error.

Worked example: use tangent

A right-angled triangle has angle $67^\circ$ and adjacent side $4.8$ cm. Find its opposite side $y$.

$$\begin{aligned} \tan67^\circ&=\frac{y}{4.8}\\ y&=4.8\tan67^\circ\\ &=11.3\text{ cm}\quad\text{to 3 s.f.} \end{aligned}$$

An angle larger than $45^\circ$ has opposite side longer than adjacent side, so $y>4.8$ is sensible.

Finding an angle with an inverse function

If a side ratio is known, use an inverse trigonometric function:

$$\theta=\sin^{-1}\left(\frac OH\right),\qquad \theta=\cos^{-1}\left(\frac AH\right),\qquad \theta=\tan^{-1}\left(\frac OA\right).$$

Here $\sin^{-1}$ means inverse sine, not $1/\sin\theta$. On a calculator it may appear as asin or above the sine key.

Worked example: find an angle

The side opposite $\theta$ is $7.2$ cm and the adjacent side is $10.5$ cm. Find $\theta$.

$$\begin{aligned} \tan\theta&=\frac{7.2}{10.5}\\ \theta&=\tan^{-1}\left(\frac{7.2}{10.5}\right)\\ &=34.4^\circ\quad\text{to 1 d.p.} \end{aligned}$$

Keep the fraction unrounded inside the inverse function. Rounding an intermediate ratio can reduce the accuracy of the final angle.

Degree mode matters

At this stage, angles are usually measured in degrees. Check that your calculator displays DEG. For example,

$\sin30^\circ=0.5.$

If your calculator does not return $0.5$, it is probably in radian mode. Radians become essential later, but the calculator mode must always match the unit of the angle.

Exact trigonometric values

The values for $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$ and $90^\circ$ should be known exactly.

$\theta$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
$\sin\theta$	$0$	$\frac12$	$\frac{\sqrt2}{2}$	$\frac{\sqrt3}{2}$	$1$
$\cos\theta$	$1$	$\frac{\sqrt3}{2}$	$\frac{\sqrt2}{2}$	$\frac12$	$0$
$\tan\theta$	$0$	$\frac1{\sqrt3}$	$1$	$\sqrt3$	undefined

The sine row rises from $0$ to $1$. The cosine row is the sine row in reverse because

$\cos\theta=\sin(90^\circ-\theta).$

Also,

$\tan\theta=\frac{\sin\theta}{\cos\theta}.$

This explains why $\tan90^\circ$ is undefined: it would require division by zero.

Where the exact values come from

Bisect an equilateral triangle of side $2$. Each half is a $30^\circ$-$60^\circ$-$90^\circ$ triangle with hypotenuse $2$, shorter side $1$ and remaining side

$\sqrt{2^2-1^2}=\sqrt3.$

Its side ratios give the exact values at $30^\circ$ and $60^\circ$.

Cut a square of side $1$ along a diagonal. Pythagoras gives diagonal $\sqrt2$, so the resulting $45^\circ$-$45^\circ$-$90^\circ$ triangle gives

$\sin45^\circ=\cos45^\circ=\frac1{\sqrt2}=\frac{\sqrt2}{2}.$

Worked example: an exact length

A right-angled triangle has hypotenuse $10$ cm and an acute angle of $60^\circ$. Find the adjacent side exactly.

$$\begin{aligned} \cos60^\circ&=\frac{x}{10}\\ \frac12&=\frac{x}{10}\\ x&=5\text{ cm}. \end{aligned}$$

When the angle has a standard exact value, do not replace it with a rounded decimal.

Combining Pythagoras and trigonometry

Some problems require more than one stage. Draw and label the triangle, decide which missing quantity unlocks the next step, and keep full calculator accuracy until the final line.

Worked example: diagonal and angle of a rectangle

A rectangle is $12$ cm long and $5$ cm wide. Find the length of its diagonal and the angle $\theta$ between the diagonal and the longer side.

The diagonal is the hypotenuse:

$$d=\sqrt{12^2+5^2}=\sqrt{169}=13\text{ cm}.$$

For the angle at the end of the $12$ cm side,

$$\begin{aligned} \tan\theta&=\frac5{12}\\ \theta&=\tan^{-1}\left(\frac5{12}\right)\\ &=22.6^\circ\quad\text{to 1 d.p.} \end{aligned}$$

The other acute angle is $90^\circ-22.6^\circ=67.4^\circ$.

Worked example: coordinate distance

Find the distance between $A(-2,3)$ and $B(4,11)$.

The horizontal and vertical changes are

$\Delta x=4-(-2)=6,\qquad \Delta y=11-3=8.$

These form the shorter sides of a right-angled triangle, so

$AB=\sqrt{6^2+8^2}=10.$

This is the source of the coordinate distance formula

$\boxed{d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}.$

The signs of the coordinate changes do not affect the distance after squaring, but calculating each difference consistently avoids mistakes.

Modelling with angles of elevation and depression

An angle of elevation is measured upwards from a horizontal line. An angle of depression is measured downwards from a horizontal line. Horizontal lines are parallel, so an angle of depression often equals the alternate angle of elevation in the triangle.

Worked example: height including eye level

A student stands $28$ m from the base of a vertical tower. The angle of elevation from eye level to the top is $41^\circ$. The student’s eye level is $1.6$ m above the ground. Find the tower’s height.

Let $x$ be the vertical rise from eye level to the top. Then

$$\begin{aligned} \tan41^\circ&=\frac{x}{28}\\ x&=28\tan41^\circ\\ &\approx24.340. \end{aligned}$$

The tower’s full height is

$24.340+1.6=25.9\text{ m}\quad\text{to 3 s.f.}$

The trigonometric calculation found the height above the observer’s eye, not the height above the ground. Translating the context into the correct triangle is part of the mathematics.

Common misconceptions

Treating any triangle as right-angled

Pythagoras and SOH CAH TOA require a right angle. For a non-right-angled triangle, use the sine rule, cosine rule or triangle area formula, or split the shape into right-angled triangles if justified.

Calling the sloping side the hypotenuse

Orientation is irrelevant. The hypotenuse is opposite the right angle, even if it is drawn horizontally or vertically.

Choosing opposite and adjacent from the diagram alone

These labels depend on the chosen angle. Circle the angle before labelling the sides.

Rounding too early

Store intermediate values or keep exact expressions such as $\sqrt{170}$. Round only the requested final answer, usually to a stated number of significant figures or decimal places.

Giving an impossible answer

Check that the hypotenuse is longest, all lengths are positive, and both acute angles lie between $0^\circ$ and $90^\circ$.

Self-check

1. A right-angled triangle has shorter sides $9$ cm and $12$ cm. Find its hypotenuse.
2. A right-angled triangle has hypotenuse $20$ cm and one shorter side $16$ cm. Find the other side.
3. Relative to an angle $\theta$, the opposite side is $6$ and the hypotenuse is $10$. Find $\theta$ to one decimal place.
4. A ladder of length $5.5$ m makes an angle of $68^\circ$ with level ground. How high up a vertical wall does it reach? Give your answer to three significant figures.
5. Find the exact value of $\sin60^\circ\cos45^\circ$.
6. A cuboid has side lengths $3$ cm, $4$ cm and $12$ cm. Find the length of the diagonal joining opposite vertices.
7. Explain why a triangle with side lengths $7$, $24$ and $25$ is right-angled.

Answers

1. $\sqrt{9^2+12^2}=15$ cm.
2. $\sqrt{20^2-16^2}=12$ cm.
3. $\theta=\sin^{-1}(6/10)=36.9^\circ$.
4. $5.5\sin68^\circ=5.10$ m.
5. $\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt2}{2}=\dfrac{\sqrt6}{4}$.
6. First find a face diagonal: $\sqrt{3^2+4^2}=5$. Then the space diagonal is $\sqrt{5^2+12^2}=13$ cm. Equivalently, $\sqrt{3^2+4^2+12^2}=13$ cm.
7. The longest side is $25$, and $7^2+24^2=49+576=625=25^2$.

What to learn next

You should now be able to choose between Pythagoras and a trigonometric ratio, calculate missing sides and angles, and interpret answers in context.

Next, study:

• the sine rule, cosine rule and triangle area for non-right-angled triangles;
• coordinate geometry for distances, midpoints and gradients;
• circle geometry for right angles in semicircles and geometric reasoning;
• exact arithmetic to manipulate exact trigonometric values confidently.