Finding missing sides using trigonometry — Mathematics

Before this

SOH CAH TOA — sine, cosine and tangent
Rearranging formulae · coming soon
Finding shorter sides using Pythagoras · coming soon

This leads to

A ramp that comes out five centimetres long

A ramp must rise 0.6 m, from a path up to a doorway. It must make an angle of only 5° with the level ground. How long must the ramp be?

The ramp is drawn to scale. It rises 0.6 m and makes 5° with the ground. A gentle slope needs a long ramp.

The last lecture gave you the right tool. The rise is opposite the 5° angle. The ramp itself is the hypotenuse. Opposite and hypotenuse means sine. In the last lecture, every answer came from multiplying a length by a ratio, such as $5 \times sin 7 0^{\circ}$ for the ladder. So try the same move here:

0.6 \times sin 5^{\circ} \approx 0.6 \times 0.0872 \approx 0.0523 m

That says the ramp is about 5 cm long. Look at the picture. The ramp is clearly much longer than its rise, and a 5 cm ramp cannot climb 60 cm. Something has gone wrong, and the calculator did nothing wrong.

The mistake is in the move, not in the ratio. "Multiply by the ratio" worked for the ladder because of where the unknown sat in the equation. Here the unknown sits somewhere else. This lecture is about that one question: where is the unknown? Once you can answer it, you can find any missing side of a right-angled triangle from one side and one angle.

One equation, two places for the unknown

The last lecture ended with three equations and a promise. Each equation had one unknown, and the promise was to solve equations like them, "including the kind where the unknown is on the bottom of the fraction". Here is that idea.

Every trigonometry equation has the same shape:

ratio = \frac{top side}{bottom side}

One of the two sides is known and the other is unknown. So there are only two cases.

Case 1: the unknown is on top. For the ladder, $sin 7 0^{\circ} = \frac{h}{5}$ . The unknown $h$ is divided by 5. To undo a division by 5, multiply both sides by 5:

h = 5 \times sin 7 0^{\circ}

Case 2: the unknown is on the bottom. For the ramp, $sin 5^{\circ} = \frac{0.6}{L}$ , where $L$ is the length of the ramp. Now 0.6 is divided by the unknown. This needs two moves. First, multiply both sides by $L$ , to bring it off the bottom:

L \times sin 5^{\circ} = 0.6

Now $L$ is multiplied by $sin 5^{\circ}$ . To undo that, divide both sides by $sin 5^{\circ}$ :

L = \frac{0.6}{sin 5 ^{\circ}} \approx \frac{0.6}{0.08716} \approx 6.88 m

A ramp about 6.88 m long. That matches the picture. Look at what happened in Case 2: the unknown and the ratio swapped places. The known side stayed on top.

That is it. That is the whole idea. If the unknown is on top, multiply. If the unknown is on the bottom, divide the known side by the ratio. You never need to remember which sides go with multiplying. The position of the unknown in the equation decides it.

Finding a missing side

Write the equation $ratio = \frac{top}{bottom}$ for the pair of sides you have.

If the unknown is on top: $unknown = known \times ratio$ .

If the unknown is on the bottom: $unknown = \frac{known}{ratio}$ .

If the second case still feels strange, test it with plain numbers. Suppose $3 = \frac{12}{x}$ . You know that $x = 4$ , because 12 divided by 4 is 3. You find it as $x = 12 \div 3$ , not $12 \times 3$ . A trigonometry equation works in exactly the same way. The only difference is that the number on the left is a ratio such as $sin 5^{\circ}$ , and you get it from the calculator.

Now commit to an answer before you read on.

Predict — which line gives the hypotenuse?

A right-angled triangle has an angle of 41°. The side opposite it is 7 cm. You write $sin 4 1^{\circ} = \frac{7}{H}$ . Which line gives the hypotenuse $H$ ? Choose one, then read the explanation.

This box shows the equation $sin 4 1^{\circ} = \frac{7}{H}$ and the right way to solve it.

The unknown $H$ is on the bottom. Multiply both sides by $H$ : $H sin 4 1^{\circ} = 7$ . Then divide both sides by $sin 4 1^{\circ}$ : $H = 7 \div sin 4 1^{\circ} \approx 7 \div 0.6561 \approx 10.7$ cm. Multiplying instead, $7 \times sin 4 1^{\circ} \approx 4.59$ cm, gives a hypotenuse shorter than the 7 cm side, which is impossible.

The two wrong answers fail a simple test of size, and that test is worth a section of its own. First, here is the whole method in order.

The method, with a reason for each step

1. Mark the angle and name the sides. Name the hypotenuse first, across from the right angle. Then name the opposite and the adjacent from the angle you are using. This is because a side is only "opposite" from a particular angle.
2. Find the pair. Write down the side you know and the side you want. The third side plays no part.
3. Choose the ratio for that pair. Opposite and hypotenuse means sine. Adjacent and hypotenuse means cosine. Opposite and adjacent means tangent.
4. Write the equation and put in the numbers. For example, $sin 4 1^{\circ} = \frac{7}{H}$ . Writing it out shows you where the unknown is.
5. Rearrange. Unknown on top: multiply. Unknown on the bottom: divide the known side by the ratio.
6. Calculate, round, and check. Keep the full calculator value until the last line. Then round to 3 significant figures (3 s.f.), and test the answer against the size rules below.

Step 4 is the step that students skip. It looks slow. But without it, step 5 becomes a guess between multiplying and dividing, and that guess is wrong about half the time.

Estimate before you trust the answer

A calculator never tells you that you divided when you should have multiplied. A rough estimate does. Three facts from the last lecture are enough for most checks.

First, the hypotenuse is the longest side. So a hypotenuse must come out longer than both of the other sides. Any other side must come out shorter than the hypotenuse.

Second, sine and cosine are less than 1. Multiplying by a number less than 1 makes a length smaller. Dividing by a number less than 1 makes a length bigger. This is why the hypotenuse comes from dividing by sine or cosine: it must be the biggest.

Third, a few rough values help: $sin 3 0^{\circ} = 0.5$ , $sin 1 0^{\circ} \approx 0.17$ , and $tan 4 5^{\circ} = 1$ . Below 45°, the tangent is less than 1, so the opposite is shorter than the adjacent. Above 45°, the opposite is the longer one.

Use them like this. If the angle is 10° and the hypotenuse is 20 cm, the opposite side should be about $0.17 \times 20 \approx 3.4$ cm, much shorter than 20 cm. If your working gives 115 cm for that side, it is longer than the hypotenuse, so you have divided by mistake. You can see the error before you check a single line.

Stop and think

In the ramp, $sin 5^{\circ} \approx 0.087$ . Without working it out, should the ramp be a little longer than 0.6 m, or many times longer? Why?

Show the answer

Many times longer. The ramp is the hypotenuse, and $sin 5^{\circ}$ says the rise is only about 0.087 of the ramp's length, less than one tenth. So the ramp must be more than ten times the rise: more than 6 m. The exact answer, 6.88 m, agrees. A small angle always needs a long hypotenuse for a given rise.

Worked examples

Worked Example 1 — The three equations from the last lecture

Solve the three equations set up at the end of the last lecture. (a) $sin 2 5^{\circ} = \frac{O}{15}$ . (b) $cos 5 0^{\circ} = \frac{A}{10}$ . (c) $tan 4 0^{\circ} = \frac{O}{8}$ . All lengths are in cm.

Think first. In all three, the unknown is on top. So each one is a multiplication.

(a) $O = 15 \times sin 2 5^{\circ} \approx 15 \times 0.4226 \approx 6.34$ cm (3 s.f.).

(b) $A = 10 \times cos 5 0^{\circ} \approx 10 \times 0.6428 \approx 6.43$ cm.

Check. In (a) and (b), each answer is shorter than its hypotenuse, 15 cm and 10 cm. ✓ In (c), the angle 40° is below 45°, so the opposite should be a little shorter than the adjacent, 8 cm. It is. ✓

Worked Example 2 — The unknown on the bottom

A right-angled triangle has an angle of 41°. The side opposite this angle is 7 cm. Find the hypotenuse.

Think first. The pair is opposite and hypotenuse, so the ratio is sine. The hypotenuse is the one we want, and it goes on the bottom of the fraction.

sin 4 1^{\circ} = \frac{7}{H}

Multiply both sides by $H$ : $H sin 4 1^{\circ} = 7$ . Divide both sides by $sin 4 1^{\circ}$ :

H = \frac{7}{sin 4 1 ^{\circ}} \approx \frac{7}{0.6561} \approx 10.7 cm (3 s.f.)

Check. 10.7 cm is longer than 7 cm, as a hypotenuse must be. ✓ For a second check, find the third side by Pythagoras: $10.6 7^{2} - 7^{2} \approx 8.05$ cm. Then $\frac{7}{10.67} \approx 0.656$ , which is $sin 4 1^{\circ}$ . ✓

Worked Example 3 — Tangent, with two routes

In triangle PQR, the right angle is at Q. Angle P is 63°, and QR = 9 cm. Find PQ.

From P, QR is the opposite and PQ is the adjacent. From R, the two swap: PQ is the opposite and QR is the adjacent.

Think first. From the 63° angle at P, QR does not touch P, so it is the opposite. PQ touches P and is not the hypotenuse, so it is the adjacent. Opposite and adjacent means tangent.

Route 1. $tan 6 3^{\circ} = \frac{9}{PQ}$ . The unknown is on the bottom, so divide:

PQ = \frac{9}{tan 6 3 ^{\circ}} \approx \frac{9}{1.9626} \approx 4.59 cm (3 s.f.)

Route 2. The angles of a triangle add up to 180°, so angle R is $18 0^{\circ} - 9 0^{\circ} - 6 3^{\circ} = 2 7^{\circ}$ . Label from R instead. Now PQ is the opposite and QR is the adjacent. So $tan 2 7^{\circ} = \frac{PQ}{9}$ , with the unknown on top:

PQ = 9 \times tan 2 7^{\circ} \approx 9 \times 0.5095 \approx 4.59 cm

Check. The two routes agree. ✓ Also, 63° is above 45°, so the opposite (9 cm) should be longer than the adjacent. It is about twice as long, as $tan 6 3^{\circ} \approx 1.96$ says. ✓

Route 2 shows a useful choice. You can always use the other angle to put the unknown on top. Both routes are correct, so use whichever you find clearer.

The working below makes one mistake that the size rules catch at once. Find it.

Spot the error — finding the adjacent side

A right-angled triangle has an angle of 35°. The side opposite it is 5 cm. Find the adjacent side. One line of this working is wrong. Tap the line you think is wrong. Every line has an explanation.

The working for finding the adjacent side when the angle is 35° and the opposite is 5 cm, with the mistake marked and corrected:

Line 1. The known side is the opposite, 5 cm, and the wanted side is the adjacent, A. (Correct.)
Line 2. Opposite and adjacent, so use tangent. (Correct.)
Line 3. tan 35° = 5 ÷ A. (Correct.)
Line 4. A = 5 × tan 35°. The mistake: A is on the bottom, so the known side is divided by the ratio: A = 5 ÷ tan 35°.
Line 5. A ≈ 5 × 0.7002 ≈ 3.50 cm. (This follows from the mistake. The correct answer is 5 ÷ 0.7002 ≈ 7.14 cm.)

The size rule found the error without any recalculation. When the angle is below 45°, the adjacent is the longer of the two shorter sides, so 3.50 cm could not be right.

Worked Example 4 — The ramp, completed

Return to the ramp. It rises 0.6 m and makes 5° with the ground. Find its length, and how far it runs along the ground.

Think first. From the 5° angle, the rise is the opposite. The ramp is the hypotenuse. The run along the ground is the adjacent.

Length: $sin 5^{\circ} = \frac{0.6}{L}$ , so $L = \frac{0.6}{sin 5 ^{\circ}} \approx \frac{0.6}{0.08716} \approx 6.88$ m (3 s.f.).

Run: use the two sides you were given, the rise and the angle, rather than the length you have just rounded. $tan 5^{\circ} = \frac{0.6}{r}$ , so $r = \frac{0.6}{tan 5 ^{\circ}} \approx \frac{0.6}{0.08749} \approx 6.86$ m.

Check. The run is a little shorter than the ramp, as it must be, because the ramp is the hypotenuse. ✓ The two are close because the angle is small. At 5°, the ramp is almost flat.

Worked Example 5 — A sentence becomes a triangle

A ramp in a loading area is 20 m long and makes an angle of 18° with the level ground. How high does it rise, and how far does it run along the ground?

Think first. Draw the triangle and mark the 18° angle at the bottom. The ramp, 20 m, is the hypotenuse. The rise is the opposite and the run is the adjacent. Both unknowns are on top.

Rise: $20 \times sin 1 8^{\circ} \approx 20 \times 0.3090 \approx 6.18$ m (3 s.f.).

Run: $20 \times cos 1 8^{\circ} \approx 20 \times 0.9511 \approx 19.0$ m (3 s.f.).

Check. Pythagoras: $6.18 0^{2} + 19.0 2^{2} \approx 38.2 + 361.8 = 400.0$ , and $2 0^{2} = 400$ . ✓ The rise is about 0.3 of the ramp, which fits an angle between 10° and 30°. ✓

Real problems arrive as sentences. The trigonometry starts only after you have drawn the triangle and marked the angle. The rest is the same six steps.

Three traps

Trap 1 — Multiplying when the unknown is on the bottom

This is the ramp mistake: $0.6 \times sin 5^{\circ}$ instead of $0.6 \div sin 5^{\circ}$ . Always write the equation first, and look at where the unknown sits. Then check the size: a hypotenuse that comes out shorter than another side means you multiplied when you should have divided.

Trap 2 — Rounding the ratio too early

If you round $sin 4 1^{\circ}$ to 0.66 and then divide, you get $7 \div 0.66 \approx 10.6$ cm. The correct answer is 10.7 cm. Dividing by a rounded number spreads its error into the answer. Keep the full calculator value, or at least 4 decimal places, until the last line.

Trap 3 — Turning the whole fraction upside down

From $sin 4 1^{\circ} = \frac{7}{H}$ , some students write $H = \frac{sin 4 1 ^{\circ}}{7}$ . That gives about 0.094 cm, less than 1 mm. The known side always stays on top: $H = \frac{7}{sin 4 1 ^{\circ}}$ . Only the ratio and the unknown change places.

Try it yourself

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

A kite string is 25 m long and makes an angle of 32° with the level ground at your hand. How far along the ground is the point directly under the kite? Give your answer to 3 s.f.

Hint 1
The string is the hypotenuse. Does the distance along the ground touch the 32° angle?

Hint 2
The ground distance is the adjacent. Adjacent and hypotenuse means cosine: $cos 3 2^{\circ} = \frac{d}{25}$ .

Full answer
The unknown is on top, so multiply: $d = 25 \times cos 3 2^{\circ} \approx 25 \times 0.8480 \approx 21.2$ m (3 s.f.). Check: 21.2 m is shorter than the 25 m string. ✓ The height of the kite is $25 \times sin 3 2^{\circ} \approx 13.2$ m, and $21.2 0^{2} + 13.2 5^{2} \approx 449.5 + 175.5 = 625 = 2 5^{2}$ . ✓
A rope holds up a straight pole. The rope is tied to the top of the pole and pegged to the level ground 6.5 m from the foot of the pole. It makes an angle of 58° with the ground. How long is the rope?

Hint 1
The rope is the hypotenuse, and it is the unknown. The 6.5 m along the ground touches the 58° angle.

Hint 2
$cos 5 8^{\circ} = \frac{6.5}{R}$ . Where is the unknown?

Full answer
The unknown is on the bottom, so divide: $R = \frac{6.5}{cos 5 8 ^{\circ}} \approx \frac{6.5}{0.5299} \approx 12.3$ m (3 s.f.). Check: the rope is longer than 6.5 m, as a hypotenuse must be. ✓ Estimate: $cos 5 8^{\circ}$ is a little more than 0.5, so the rope should be a little less than twice 6.5 m, which is 13 m. ✓
A right-angled triangle has an angle of 22°. The side opposite it is 4.2 cm. Find the adjacent side in two ways: once with the 22° angle, and once with the other angle. Then use a size rule to check.

Hint 1
With the 22° angle, $tan 2 2^{\circ} = \frac{4.2}{A}$ . The other angle is $9 0^{\circ} - 2 2^{\circ}$ .

Hint 2
From the other angle, 68°, the 4.2 cm side touches the angle, so it is the adjacent, and the side you want is the opposite.

Full answer
With 22°: the unknown is on the bottom, so $A = \frac{4.2}{tan 2 2 ^{\circ}} \approx \frac{4.2}{0.4040} \approx 10.4$ cm (3 s.f.). With 68°: $tan 6 8^{\circ} = \frac{x}{4.2}$ , so $x = 4.2 \times tan 6 8^{\circ} \approx 4.2 \times 2.4751 \approx 10.4$ cm. ✓ Check: 22° is below 45°, so the side opposite it (4.2 cm) must be shorter than the adjacent. 10.4 cm is longer. ✓ The wrong move, $4.2 \times tan 2 2^{\circ} \approx 1.70$ cm, fails this test.
A stretch. A triangle has two equal sides, each 10 cm long. The angles at the ends of the third side are both 40°. Find the length of the third side, the height of the triangle, and its area.

Hint 1
This triangle has no right angle. Draw the line from the top corner straight down to the third side. It splits the triangle into two equal right-angled triangles.

Hint 2
In each half, the hypotenuse is 10 cm and the angle at the bottom is 40°. The height is the opposite. Half of the third side is the adjacent.

Full answer
The line down from the top meets the third side at a right angle and cuts it in half, because the triangle is symmetric. In each half: height $h = 10 \times sin 4 0^{\circ} \approx 10 \times 0.6428 \approx 6.43$ cm, and half the third side is $10 \times cos 4 0^{\circ} \approx 10 \times 0.7660 \approx 7.66$ cm. So the third side is about $2 \times 7.660 \approx 15.3$ cm (3 s.f.). Area $= \frac{1}{2} \times base \times height \approx \frac{1}{2} \times 15.32 \times 6.428 \approx 49.2$ cm². Check: in each half, $7.66 0^{2} + 6.42 8^{2} \approx 58.68 + 41.32 = 100 = 1 0^{2}$ . ✓ Splitting a shape into right-angled triangles is how trigonometry reaches triangles that have no right angle. A later lecture turns this idea into a single formula for the area.

The bigger idea

Trigonometry lets you find a length you cannot reach from one length you can. On 10 April 1802, the surveyor William Lambton began the Great Trigonometrical Survey of India with a single measured line near St Thomas Mount, south of Madras. From that one line, the survey built a chain of huge triangles across the country. Each new side was calculated, not walked, from angles and sides already known. It took about forty years to reach the Himalayas.

The same method gave the heights of mountains that nobody could climb. In 1852, Radhanath Sikdar, the survey's chief computer, worked through the observations of a peak then called Peak XV. In 1856 the survey announced it as probably the highest mountain in the world, at 29 002 feet (about 8840 m). It is now called Mount Everest. (Sources: Royal Geographical Society, notes on the Great Trigonometrical Survey; S. Mishra et al., Indian Journal of History of Science, 2015.) Those calculations used more than one right-angled triangle, but the move at their heart is the one in this lecture: one side, one angle, and the right equation.

Key concepts in this lesson

To find a missing side, name the sides from the angle, choose the ratio for the pair of sides you know and want, and write the equation before you rearrange.
If the unknown is on top of the fraction, multiply the known side by the ratio.
If the unknown is on the bottom, divide the known side by the ratio, so the unknown and the ratio change places.
Using the other angle, $9 0^{\circ} - θ$ , can put the unknown on top, and both routes give the same answer.
Check every answer by size: the hypotenuse is the longest side, and below 45° the opposite is shorter than the adjacent.