SOH CAH TOA — sin cos tan introduced — Mathematics

Before this

Similar triangles · coming soon
Pythagoras' theorem · coming soon
Gradient · coming soon

This leads to

A ladder that Pythagoras cannot measure

A ladder 5 m long leans against a wall. Its foot stands on level ground, and the ladder makes an angle of 70° with the ground. How high up the wall does it reach?

A 5 m ladder makes 70° with level ground. How high up the wall does it reach? You know one side and one angle, not two sides.

The ladder, the wall and the ground make a right-angled triangle, so your first tool is Pythagoras' theorem. But Pythagoras needs two sides to find the third, and here you know only one. What you know instead is an angle, and Pythagoras' theorem has no place to put an angle.

So you need a new idea, one that connects the angles of a triangle to the lengths of its sides. That connection is called trigonometry, from Greek words meaning "triangle measuring". Its three basic tools are sine, cosine and tangent. Why they work matters more than their names, so we will build the idea one move at a time.

Draw a small copy

Here is a way to answer with a ruler and a protractor. On paper, draw a right-angled triangle with an angle of 70° and a sloping side of exactly 10 cm. Measure its vertical side. If you draw carefully, you get about 9.4 cm.

Your small triangle has the same angles as the ladder triangle, so it has the same shape. The ladder triangle is your paper triangle, made bigger. In your drawing, the vertical side is 0.94 of the slope, and making the triangle bigger does not change this fraction. So the height up the wall is also 0.94 of the ladder:

0.94 \times 5 = 4.7 m

Look at what just happened there. Nobody measured the wall. A small drawing gave a number, 0.94, and that number belongs to the angle 70°, in every right-angled triangle, whatever its size.

Trigonometry is this trick made exact. The fractions have been worked out precisely for every angle, and your calculator gives them in a moment. The rest of this lecture explains why the trick is allowed, and gives the fractions their names.

Same angle, same shape, same ratios

The trick rests on one fact. Take any right-angled triangle, and choose one of its other two angles. Call it $θ$ . This is the Greek letter theta, the usual letter for an angle.

The three angles of a triangle add up to 180°. One angle is 90°, and one is $θ$ . So the third angle must be $18 0^{\circ} - 9 0^{\circ} - θ = 9 0^{\circ} - θ$ . This means that once you choose $θ$ , all three angles are fixed. Every right-angled triangle with an angle $θ$ has exactly the same three angles.

Triangles with the same three angles are similar: one is an enlargement of the other. In an enlargement, every length is multiplied by the same scale factor, $k$ .

Now watch what happens to a ratio of two sides. Suppose two sides of the small triangle have lengths $p$ and $q$ . In the big triangle, they become $k p$ and $k q$ . The ratio of the big pair is

\frac{k p}{k q} = \frac{p}{q} .

The $k$ cancels. So the ratio is the same in both triangles. The lengths change, but the ratios do not.

That is it. That is the whole idea. In a right-angled triangle, the ratio of any two sides depends only on the angle, not on the size of the triangle. Everything else in this lecture is a name for one of these ratios, or a way to use them.

Three right-angled triangles with the same 35° angle. Each is an enlargement of the others, so they have the same shape.

The three sloping sides are 4 cm, 6 cm and 8 cm, and the vertical sides are about 2.29 cm, 3.44 cm and 4.59 cm. Divide each vertical side by its sloping side: every answer is about 0.574.

Stop and think

A student says: "If I make the triangle twice as big, the vertical side doubles. So the ratio vertical ÷ sloping doubles too." What is wrong with this?

Show the answer

The sloping side doubles too, and $\frac{2 p}{2 q} = \frac{p}{q}$ . For example, 2.29 ÷ 4 and 4.59 ÷ 8 are both about 0.574. Only a change in the angle changes the ratio.

Test this in the explorer below. One control changes the size of the triangle, and the other changes its angle.

Explore — change the size, then change the angle

First move the size slider and watch the three ratios. Then move the angle slider and watch them again.

Hypotenuse H: 8 cm

Angle:

Angle θ: 35°

This table shows right-angled triangles with three different angles, each drawn in two sizes. O is the opposite, A the adjacent and H the hypotenuse, in cm, rounded to 2 decimal places. The ratios are rounded to 3 decimal places.

Angle	H	O	A	O ÷ H	A ÷ H	O ÷ A
30°	4	2.00	3.46	0.500	0.866	0.577
30°	8	4.00	6.93	0.500	0.866	0.577
45°	4	2.83	2.83	0.707	0.707	1.000
45°	8	5.66	5.66	0.707	0.707	1.000
60°	4	3.46	2.00	0.866	0.500	1.732
60°	8	6.93	4.00	0.866	0.500	1.732

Doubling the size doubles every length but leaves every ratio unchanged. Changing the angle changes the ratios.

You should have seen two different behaviours. Changing the size changed all three lengths but none of the ratios. Changing the angle changed the ratios. A ratio is a fingerprint of an angle.

Naming the sides from the angle

Words like "vertical" and "bottom" break as soon as a triangle is turned around. So the sides are named from the angle you are working with instead.

Start with the side that never changes its name. The hypotenuse is the side across from the right angle. It is always the longest side. Name it first, every time.

Then look at your angle $θ$ . Two sides remain. The opposite is the side across from $θ$ : it does not touch $θ$ at all. The adjacent is the side that touches $θ$ and is not the hypotenuse. "Adjacent" means "next to". It runs from $θ$ to the right angle.

The same triangle labelled twice. In the first picture, from θ, the opposite is the right-hand side. In the second, from φ, the opposite is the bottom side. The hypotenuse never changes.

The two pictures show the same triangle. In the second, the marked angle is the other one, $φ$ (the Greek letter phi), at the top. The hypotenuse stays the same, but the opposite and adjacent swap places. A side is not "the opposite" by itself. It is the opposite of a particular angle.

The hypotenuse also touches $θ$ . That is why you name it first: then "adjacent" means the other side that touches the angle.

Three ratios, three names

A right-angled triangle has three sides, so there are three ways to choose a pair of them: opposite with hypotenuse, adjacent with hypotenuse, and opposite with adjacent. Each pair gives one ratio, and each ratio has a name.

Sine, cosine and tangent

In a right-angled triangle, for an angle $θ$ that is not the right angle:

$sin θ = \frac{opposite}{hypotenuse} cos θ = \frac{adjacent}{hypotenuse} tan θ = \frac{opposite}{adjacent}$

Each one depends only on $θ$ , not on the size of the triangle.

Notice that $sin 3 5^{\circ}$ is not a length and not an angle. It is a pure number, about 0.574, that belongs to the angle 35°, in every triangle you will ever meet.

Many students hold the definitions with the word SOH CAH TOA: Sine = Opposite ÷ Hypotenuse, Cosine = Adjacent ÷ Hypotenuse, Tangent = Opposite ÷ Adjacent. The word only pairs each name with its sides. The reason the ratios exist at all is the similar triangles.

The real skill is choosing the ratio. Find the two sides in the question, the one you know and the one you want, and choose the ratio that uses that pair. Practise on the cards below.

Sort — sine, cosine or tangent?

Each card names the two sides in a problem: one you know and one you want. Choose a card, then choose the ratio that uses exactly that pair.

This table matches each pair of sides to the ratio that uses it.

The two sides in the problem	Ratio
You know the hypotenuse and want the side opposite θ.	sine (O and H)
You know the side opposite θ and want the hypotenuse.	sine (O and H)
You know the adjacent side and want the hypotenuse.	cosine (A and H)
You know the side opposite θ and want the adjacent side.	tangent (O and A)
A ladder of known length; you want the height it reaches, measured from the angle at its foot.	sine (O and H)
A kite string of known length; you want the distance along the ground, measured from the angle at your hand.	cosine (A and H)
A ramp: you know how far it runs along the ground and want how high it rises, from the angle at the bottom.	tangent (O and A)
The angle is at the top of the triangle. You know the hypotenuse and want the side that touches that top angle.	cosine (A and H)

The card with the angle at the top is the one to read twice: always label from the angle you are using.

What the numbers can be, and what they cannot be

Sine and cosine are always less than 1. The hypotenuse is the longest side. So opposite ÷ hypotenuse and adjacent ÷ hypotenuse divide a smaller number by a bigger one. If a calculation ever gives $sin θ = 1.2$ , something has gone wrong.

Tangent can be any positive size. At 45°, the triangle is isosceles, so the opposite equals the adjacent and $tan 4 5^{\circ} = 1$ . Above 45°, the opposite is the longer of the two, so $tan θ$ is bigger than 1. Near 90°, the adjacent becomes tiny: $tan 8 5^{\circ} \approx 11.4$ .

The "co" in cosine has a meaning. Look again at the two labelled triangles. The side opposite $θ$ is the side adjacent to the other angle, $9 0^{\circ} - θ$ . So the cosine of one angle is the sine of the other:

cos θ = sin (9 0^{\circ} - θ)

Two angles that add up to 90° are called complementary. The cosine is the "sine of the complement", which is exactly where the name comes from. The short form "co.sinus" appears in a table of 1620 by the English mathematician Edmund Gunter. Check it: $cos 6 0^{\circ}$ and $sin 3 0^{\circ}$ are both 0.5.

Tangent is a gradient. For a road that climbs at an angle $θ$ , the rise is the opposite and the horizontal run is the adjacent. So $tan θ$ = rise ÷ run, which is exactly the gradient of the road. A ramp at 5° has a gradient of about 0.087.

Stop and think

$sin 3 0^{\circ} = 0.5$ . Is $sin 6 0^{\circ}$ equal to $2 \times 0.5 = 1$ ?

Show the answer

No. $sin 6 0^{\circ} \approx 0.866$ . When the angle doubles, the triangle changes shape, and the ratio changes in its own way. Also, a sine of 1 would make the opposite as long as the hypotenuse, which is impossible. $sin θ$ is not "sin times $θ$ ". It means the ratio that belongs to the angle $θ$ .

Getting the numbers from a calculator

A scientific calculator has keys marked sin, cos and tan. Calculators measure angles in degrees or radians, and at this level you always want degrees. Look for "D" or "DEG" on the screen. If it is not there, use the "Mode" or "Setup" key.

Then test it with three values you can trust: $sin 3 0^{\circ} = 0.5$ , $cos 6 0^{\circ} = 0.5$ and $tan 4 5^{\circ} = 1$ . Two more values show the rules above at work. $cos 4 5^{\circ} \approx 0.707$ (3 significant figures, written 3 s.f. from now on), the same as $sin 4 5^{\circ}$ , because the 45° triangle is isosceles. And $tan 6 0^{\circ} \approx 1.73$ (3 s.f.), bigger than 1, because 60° is more than 45°. A later lecture finds exact values for 30°, 45° and 60° without a calculator.

If the calculator is set to radians and you type sin 30, it shows about −0.988. It is not broken. It has read 30 as 30 radians, a completely different angle: one radian is about 57.3°, so 30 radians is about 1719°. The Pre-University lectures explain radians. Here, a negative sine is a signal to check the mode.

Worked examples

Worked Example 1 — The three ratios from three sides

Triangle ABC has a right angle at C. AC = 4 cm, BC = 3 cm and AB = 5 cm. Find $sin A$ , $cos A$ and $tan A$ . Then find the same three ratios for angle B.

Think first. Label from each angle in turn. AB is across from the right angle at C, so AB is the hypotenuse for both angles.

For angle A: the opposite is BC = 3 (it does not touch A), and the adjacent is AC = 4.

sin A = \frac{3}{5} = 0.6, cos A = \frac{4}{5} = 0.8, tan A = \frac{3}{4} = 0.75

For angle B: the opposite is AC = 4, and the adjacent is BC = 3. The two swap.

sin B = \frac{4}{5} = 0.8, cos B = \frac{3}{5} = 0.6, tan B = \frac{4}{3} \approx 1.33

Check. First, $3^{2} + 4^{2} = 25 = 5^{2}$ , so the triangle really is right-angled. Second, A and B add up to 90°, so $sin A$ should equal $cos B$ . Both are 0.6. ✓ Third, $tan B$ is bigger than 1, so B should be bigger than 45°: its opposite, 4, is longer than its adjacent, 3. ✓

Worked Example 2 — Which ratio links the sides?

For each right-angled triangle, choose the ratio and write an equation with one unknown. (a) The angle is 25°, the hypotenuse is 15 cm, and you want the side opposite the angle. (b) The angle is 50°, the hypotenuse is 10 cm, and you want the side adjacent to the angle. (c) The angle is 40°, the adjacent side is 8 cm, and you want the opposite side.

Think first. For each one, write down the two sides involved: the one you know and the one you want.

(a) Opposite and hypotenuse: that is sine. $sin 2 5^{\circ} = \frac{O}{15}$ .

(b) Adjacent and hypotenuse: that is cosine. $cos 5 0^{\circ} = \frac{A}{10}$ .

Check. Each equation has exactly one unknown. In (c), a ratio with the hypotenuse would bring in a second unknown. The next lecture solves equations like these, including the kind where the unknown is on the bottom of the fraction.

Worked Example 3 — The ladder, exactly

Return to the ladder. It is 5 m long and makes 70° with the ground. Find the height it reaches up the wall, and how far its foot is from the wall.

Think first. From the 70° angle at the foot, the ladder is the hypotenuse. The height up the wall does not touch the foot, so it is the opposite. The distance along the ground is the adjacent.

Height: opposite and hypotenuse, so sine. The height is the fraction $sin 7 0^{\circ}$ of the ladder: $5 \times sin 7 0^{\circ} \approx 5 \times 0.9397 \approx 4.70$ m (3 s.f.).

Distance: adjacent and hypotenuse, so cosine. $5 \times cos 7 0^{\circ} \approx 5 \times 0.3420 \approx 1.71$ m (3 s.f.).

Check. Pythagoras can now check trigonometry: $4.69 8^{2} + 1.71 0^{2} \approx 22.07 + 2.92 = 24.99$ , which is 25 when the rounding is removed, and $5^{2} = 25$ . ✓ The drawing at the start gave 4.7 m, which agrees.

Three traps

Trap 1 — Labelling from the wrong angle

If the angle is at the top of the triangle, the side at the bottom is its opposite. Students who label "opposite = the vertical side" out of habit swap sine and cosine. Mark the angle you are using, then label from it.

Trap 2 — Calling the hypotenuse "adjacent"

The hypotenuse touches the angle, but it is never called the adjacent. Name it first, across from the right angle. Then name the opposite and the adjacent from the two sides left.

Trap 3 — The calculator in radian mode

In radian mode, sin 30 gives about −0.988 instead of 0.5, and every answer after it is wrong while the working looks perfect. Test with $sin 3 0^{\circ} = 0.5$ before an exam.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Triangle PQR has a right angle at R. PR = 15 cm, QR = 8 cm and PQ = 17 cm. Find $sin P$ , $cos P$ and $tan P$ as fractions, and then as decimals to 3 s.f.

Hint 1
The hypotenuse is across from the right angle at R. Which side is that?

Hint 2
PQ is the hypotenuse. From P, the side that does not touch P is QR. The side that touches P and is not the hypotenuse is PR.

Full answer
The hypotenuse is PQ = 17. From P, the opposite is QR = 8 and the adjacent is PR = 15. So $sin P = \frac{8}{17} \approx 0.471$ , $cos P = \frac{15}{17} \approx 0.882$ , $tan P = \frac{8}{15} \approx 0.533$ . Check: $8^{2} + 1 5^{2} = 64 + 225 = 289 = 1 7^{2}$ , so the triangle is right-angled. Both sine and cosine are less than 1, and the tangent is less than 1 because the opposite, 8, is shorter than the adjacent, 15. ✓
Without a calculator, decide which is bigger in each pair, and give a reason with a triangle: (a) $sin 2 0^{\circ}$ or $sin 7 0^{\circ}$ ; (b) $cos 2 0^{\circ}$ or $cos 7 0^{\circ}$ .

Hint 1
Imagine a triangle with a fixed hypotenuse. What happens to the opposite as the angle grows?

Hint 2
As the angle grows, the opposite gets longer and the adjacent gets shorter, while the hypotenuse stays the same.

Full answer
(a) $sin 7 0^{\circ}$ is bigger. Keep the hypotenuse fixed and open the angle from 20° to 70°: the opposite side grows, so opposite ÷ hypotenuse grows. (b) $cos 2 0^{\circ}$ is bigger. As the angle opens, the adjacent shrinks, so adjacent ÷ hypotenuse shrinks. Check with a calculator: $sin 2 0^{\circ} \approx 0.342$ and $sin 7 0^{\circ} \approx 0.940$ ; $cos 2 0^{\circ} \approx 0.940$ and $cos 7 0^{\circ} \approx 0.342$ . ✓ The same two numbers appear twice because 20° and 70° are complementary: $cos 2 0^{\circ} = sin 7 0^{\circ}$ .
An angle $θ$ in a right-angled triangle has $sin θ = \frac{5}{13}$ . Find $cos θ$ and $tan θ$ as fractions, without finding $θ$ .

Hint 1
The ratio is the same in every triangle with this angle. So draw one convenient triangle: opposite 5 and hypotenuse 13.

Hint 2
Use Pythagoras to find the adjacent side: $1 3^{2} - 5^{2}$ .

Full answer
Draw a right-angled triangle with opposite 5 and hypotenuse 13. Then the adjacent is $1 3^{2} - 5^{2} = 169 - 25 = 144 = 12$ . So $cos θ = \frac{12}{13}$ and $tan θ = \frac{5}{12}$ . Any other triangle with the same angle is an enlargement of this one, so it gives the same ratios. Check: $\frac{5}{13} \div \frac{12}{13} = \frac{5}{12}$ , which matches $tan θ$ , because sine ÷ cosine is (opposite ÷ hypotenuse) ÷ (adjacent ÷ hypotenuse) = opposite ÷ adjacent. ✓
A stretch. For the 3, 4, 5 triangle in Worked Example 1, work out $(sin A)^{2} + (cos A)^{2}$ . Do the same for angle P in question 1. Then explain why you get the same answer for every angle in every right-angled triangle.

Hint 1
Write $(sin A)^{2} + (cos A)^{2}$ as $(\frac{O}{H})^{2} + (\frac{A}{H})^{2}$ , using letters for the sides.

Hint 2
Put the two fractions over the same denominator $H^{2}$ . What does Pythagoras say about the top?

Full answer
For angle A: $0. 6^{2} + 0. 8^{2} = 0.36 + 0.64 = 1$ . For angle P: $(\frac{8}{17})^{2} + (\frac{15}{17})^{2} = \frac{64 + 225}{289} = \frac{289}{289} = 1$ . In general, with sides O, A and H, $(\frac{O}{H})^{2} + (\frac{A}{H})^{2} = \frac{O ^{2} + A ^{2}}{H ^{2}}$ . By Pythagoras, $O^{2} + A^{2} = H^{2}$ , so the fraction is $\frac{H ^{2}}{H ^{2}} = 1$ . Check with a calculator at an angle that is not from a neat triangle: $sin 3 5^{\circ} \approx 0.5736$ and $cos 3 5^{\circ} \approx 0.8192$ , and $0.573 6^{2} + 0.819 2^{2} \approx 0.3290 + 0.6711 \approx 1.000$ . ✓ This is Pythagoras' theorem written in the language of trigonometry. It is one of the most used results in Pre-University mathematics.

The bigger idea

The word "sine" has a strange history. In 499 CE, the Indian astronomer Aryabhata wrote about the half-chord of a circle, which he called ardha-jya, and soon simply jya, the Sanskrit word for a bowstring. His book, the Aryabhatiya, contains one of the earliest surviving tables of these values. Arabic scholars took the word over as jiba. Arabic was usually written without vowels. So a Latin translator of the twelfth century, usually named as Robert of Chester in 1145, read it as jaib, a word for a fold or a bay. He wrote the Latin for a fold, sinus, and "sine" comes from that misreading.

The idea underneath has never changed: one number for each angle, worked out once and trusted everywhere. Astronomers used it to measure the sky, and surveyors to measure land they could not cross. Later lectures carry the same ratios into triangles without a right angle, and then into the waves of sound and light.

Key concepts in this lesson

All right-angled triangles with the same angle $θ$ are similar, so the ratio of any two of their sides depends only on $θ$ .
The hypotenuse is across from the right angle, the opposite is across from $θ$ , and the adjacent touches $θ$ and is not the hypotenuse.
$sin θ$ = opposite ÷ hypotenuse, $cos θ$ = adjacent ÷ hypotenuse and $tan θ$ = opposite ÷ adjacent, one ratio for each pair of sides.
Sine and cosine are always less than 1, tangent can be any positive number, and $cos θ = sin (9 0^{\circ} - θ)$ .
A calculator gives the ratios for any angle, but only in degree mode, which you can test with $sin 3 0^{\circ} = 0.5$ .