Maths · 7 Trigonometry

7.2 Right-angled triangles

Before you start

Pythagoras' theorem and identifying the hypotenuse (7.1).
Rearranging an equation to make any letter the subject (2.5).
Using the sin, cos, tan and inverse buttons on a calculator, in degrees.

Syllabus objectives covered (4)

Know and use the sine, cosine and tangent ratios for acute angles in calculations involving sides and angles of a right-angled triangle.
Solve problems in two dimensions using Pythagoras’ theorem and trigonometry.
Know that the perpendicular distance from a point to a line is the shortest distance to the line.
Carry out calculations involving angles of elevation and depression.

Part 1 of 4

Labelling the sides — do this first, every time

Trigonometry connects an angle to a ratio of two sides. Which two depends entirely on where the angle is, so labelling comes before anything else.
Mark the angle you are working with. Then label the three sides relative to that angle.
Hypotenuse — opposite the right angle. This one never changes, whichever angle you choose.
Opposite — the side directly across from your angle, not touching it.
Adjacent — the remaining side, next to your angle. It is the one touching both your angle and the right angle.
The opposite and adjacent swap over if you switch to the other acute angle. That is why you label for a specific angle, not for the triangle as a whole.
The three ratios, with the usual mnemonic SOH CAH TOA:
$\sin θ = \frac{O}{H} \cos θ = \frac{A}{H} \tan θ = \frac{O}{A}$
To choose the right one: identify which two sides the question involves — the one you know and the one you want — and pick the ratio containing exactly those two.

Part 2 of 4

Finding a side, and finding an angle

Finding a side. You know an angle and one side, and want another side.
Write the ratio, substitute, and rearrange. To find $x$ opposite a $35 °$ angle with hypotenuse $12$ cm: $\sin 35 ° = x / 12$ , so $x = 12 \sin 35 ° = 6.88$ cm.
If the unknown is in the denominator, rearrange carefully. $\cos 40 ° = 8 / x$ gives $x = 8 / \cos 40 °$ , not $8 \cos 40 °$ .
Finding an angle. You know two sides and want the angle between or beside them.
Form the ratio, then use the inverse function — $\sin^{- 1}$ , $\cos^{- 1}$ or $\tan^{- 1}$ on your calculator. If $\tan θ = 5 / 7$ then $θ = \tan^{- 1} (5 / 7) = 35.5 °$ .
The syllabus notes are specific about accuracy: angles are given and answered in degrees, with decimals correct to one decimal place. So $35.5 °$ , not $35.54 °$ and not $36 °$ .
Check the calculator is in degree mode. In radians every answer is wrong, and the numbers look plausible enough that you may not notice.
Sanity-check every answer. In a right-angled triangle the two acute angles must each be less than $90 °$ and add to $90 °$ , and no side can be longer than the hypotenuse.

Part 3 of 4

Elevation, depression and bearings

An angle of elevation is measured upwards from the horizontal — looking up at the top of a tower.
An angle of depression is measured downwards from the horizontal — looking down from a cliff at a boat.
Both are measured from the horizontal, never from the vertical. Measuring from a vertical wall gives the complement of the angle you want, and every subsequent step is wrong.
A useful fact: the angle of depression from A down to B equals the angle of elevation from B up to A, because they are alternate angles between parallel horizontals.
Depression from the top equals elevation from the bottom — alternate angles.
That equality is often the step that unlocks a question, because it moves the angle into the triangle you actually want to solve.
The syllabus notes add that knowledge of bearings may be required. A bearing is measured clockwise from north and always written with three figures — $070 °$ , not $70 °$ .
With any of these, draw the diagram before calculating. Mark the horizontal, mark the angle in the right place, and only then look for a right-angled triangle.

Part 4 of 4

Two-dimensional problems, and shortest distance

Most exam questions are not bare triangles. They are situations — a ramp, a roof, a ship's course — where you must find the right-angled triangle yourself.
The approach: sketch it, mark everything you know, look for or create a right angle, then choose Pythagoras or trigonometry.
Which to use? Pythagoras needs three sides and no angle. Trigonometry involves an angle. If a question gives two sides and asks for an angle, that is trigonometry; three sides and no angle mentioned, that is Pythagoras.
Multi-step problems are common: find a length with Pythagoras, then use it to find an angle with trigonometry. Carry the unrounded value between steps.
Shortest distance. The perpendicular distance from a point to a line is the shortest distance from that point to the line.
Any other line from the point to the line is the hypotenuse of a right-angled triangle whose short side is the perpendicular — and the hypotenuse is always longer. That is the reason, and it is worth being able to state.
In practice this tells you where to draw. 'How close does the ship pass to the lighthouse?' means drop a perpendicular from the lighthouse to the ship's course, and solve that right-angled triangle.

Worked examples

Example 1. In a right-angled triangle, one acute angle is $28 °$ and the side adjacent to it is $9.4$ cm. Calculate the length of the hypotenuse.

Show the working

Label relative to the $28 °$ angle: the $9.4$ cm side is adjacent, and the hypotenuse is what you want.
Adjacent and hypotenuse together means cosine — the C and A and H of CAH.
$\cos 28 ° = \frac{9.4}{h}$
The unknown is in the denominator, so rearrange by multiplying both sides by $h$ and dividing by $\cos 28 °$ .
$h = \frac{9.4}{\cos 28 °}$
Evaluate with the calculator in degree mode.
$h = \frac{9.4}{0.8829} = 10.6 cm$
Check: the hypotenuse must be longer than $9.4$ cm, and $10.6 > 9.4$ . Correct.

Answer: $10.6$ cm (3 s.f.).

Example 2. From a point $50$ m from the base of a tower, the angle of elevation of the top is $37 °$ . Calculate the height of the tower.

Show the working

Sketch it: the tower is vertical, the $50$ m is horizontal along the ground, and the angle of elevation is between the horizontal and the line of sight.
Label relative to the $37 °$ angle. The $50$ m ground distance is adjacent; the tower height is opposite.
Opposite and adjacent means tangent.
$\tan 37 ° = \frac{h}{50}$
Rearrange and evaluate.
$h = 50 \tan 37 ° = 50 \times 0.7536 = 37.7 m$
Sense-check: the angle is less than $45 °$ , so the height should be less than the ground distance of 50 m. It is.

Answer: The tower is $37.7$ m tall (3 s.f.).

Example 3. A right-angled triangle has an opposite side of $6.2$ cm and a hypotenuse of $9.8$ cm. Calculate the angle, giving your answer correct to one decimal place.

Show the working

The two sides involved are the opposite and the hypotenuse, so use sine.
$\sin θ = \frac{6.2}{9.8}$
Evaluate the ratio: $6.2 / 9.8 = 0.6327$ .
Use the inverse sine to recover the angle.
$θ = \sin^{- 1} (0.6327)$
The calculator gives $39.25 °$ , which to one decimal place is $39.3 °$ — the accuracy the syllabus specifies for angles.
Check the ratio is less than 1, as it must be when the opposite side is shorter than the hypotenuse. A ratio above 1 would mean the sides were mislabelled.

Answer: $39.3 °$ .

Example 4. A ship sails on a straight course. At its closest, it passes a lighthouse. When the ship is $800$ m from the lighthouse, the angle between its course and the line to the lighthouse is $34 °$ . Calculate the shortest distance between the ship's course and the lighthouse.

Show the working

The shortest distance from a point to a line is the perpendicular distance, so draw the perpendicular from the lighthouse to the ship's course.
That creates a right-angled triangle: the $800$ m line to the lighthouse is the hypotenuse, and the perpendicular distance is the side opposite the $34 °$ angle.
Opposite and hypotenuse means sine.
$\sin 34 ° = \frac{d}{800}$
Rearrange and evaluate.
$d = 800 \sin 34 ° = 800 \times 0.5592 = 447 m$
Check: the perpendicular is a short side, so it must be less than the $800$ m hypotenuse. It is.

Answer: $447$ m (3 s.f.).

Where people go wrong

Mistake: Labelling opposite and adjacent before choosing which angle to work from.

Instead: Mark the angle first, then label the sides relative to it.

Opposite and adjacent swap when you switch to the other acute angle. Only the hypotenuse is fixed.

Mistake: Writing $x = 8 \cos 40 °$ when the equation is $\cos 40 ° = 8 / x$ .

Instead: Rearranging gives $x = 8 / \cos 40 °$ .

When the unknown is in the denominator the operation reverses. Multiply both sides by $x$ first and see what happens.

Mistake: Measuring an angle of elevation from the vertical.

Instead: Both elevation and depression are measured from the horizontal.

Measuring from the vertical gives the complementary angle, so every result is wrong by the difference from $90 °$ .

Mistake: Leaving the calculator in radian mode.

Instead: Set it to degrees.

This syllabus works entirely in degrees, and radian answers are plausible-looking but wrong.

Mistake: Giving an angle to 3 significant figures.

Instead: Angles in degrees go to one decimal place.

The syllabus states this accuracy rule explicitly, and it is a separate rule from the 3 significant figures used for lengths.

Mistake: Rounding an intermediate length before using it in the next step.

Instead: Carry the full unrounded value forward.

Multi-step problems compound rounding errors, and the accuracy mark is lost even when the method is right.

Check yourself

Write down the three trigonometric ratios in terms of opposite, adjacent and hypotenuse.
Answer
$\sin θ = O / H$ , $\cos θ = A / H$ , $\tan θ = O / A$ .
Which ratio would you use to find the opposite side, given the angle and the adjacent side?
Answer
Tangent, since it relates the opposite and adjacent sides.
Calculate $x$ if $\sin 42 ° = x / 15$ .
Answer
$x = 15 \sin 42 ° = 10.0$ (3 s.f.).
Find the angle if the adjacent side is $7$ cm and the hypotenuse is $11$ cm.
Answer
$\cos θ = 7 / 11$ , so $θ = \cos^{- 1} (0.6364) = 50.5 °$ .
Explain the difference between an angle of elevation and an angle of depression.
Answer
Elevation is measured upwards from the horizontal; depression is measured downwards from the horizontal.
Why does the angle of depression from a cliff top to a boat equal the angle of elevation from the boat to the cliff top?
Answer
The two horizontals are parallel, so the angles are alternate angles and therefore equal.
Why is the perpendicular distance from a point to a line the shortest distance?
Answer
Any other line from the point forms the hypotenuse of a right-angled triangle whose short side is the perpendicular, and the hypotenuse is always longer.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A ramp rises $1.4$ m over a horizontal distance of $8.0$ m.

(a) Calculate the angle the ramp makes with the horizontal, correct to one decimal place. 3 marks

Mark scheme

M1 tan θ = 1.4 / 8.0 oe
M1 θ = tan⁻¹(0.175) seen
A1 9.9° awrt

(b) Calculate the length of the sloping surface of the ramp. 2 marks

Mark scheme

M1 √(1.4² + 8.0²) oe, or 8.0 / cos(their angle)
A1 8.12 awrt (m)

Question 2. A bird sits at the top of a vertical pole. From a point $P$ on level ground, the angle of elevation of the bird is $52 °$ . $P$ is $18$ m from the foot of the pole.

(a) Calculate the height of the bird above the ground. 3 marks

Mark scheme

M1 identifies 18 as adjacent and height as opposite
M1 h = 18 tan 52° oe
A1 23.0 awrt (m)

(b) Write down the angle of depression of $P$ from the bird, and give a reason. 2 marks No calculator

Mark scheme

B1 52°
B1 alternate angles between parallel horizontal lines oe

(c) Calculate the direct distance from $P$ to the bird. 2 marks

Mark scheme

M1 18 / cos 52° oe, or Pythagoras with their (a)
A1 29.2 awrt (m); FT their (a)

Question 3. A straight road runs east–west. A house $H$ stands $340$ m from a point $A$ on the road, and the angle between $A H$ and the road is $23 °$ .

(a) Calculate the shortest distance from the house to the road. 3 marks

Mark scheme

M1 identifies the shortest distance as the perpendicular from H to the road
M1 d = 340 sin 23° oe
A1 133 awrt (m)

(b) Explain why the perpendicular gives the shortest distance. 1 mark No calculator

Mark scheme

B1 any other route is the hypotenuse of a right-angled triangle containing the perpendicular, and the hypotenuse is longer oe

Words to know

Opposite: The side directly across from the angle being used.
Adjacent: The side next to the angle being used, touching both it and the right angle.
Angle of elevation: The angle measured upwards from the horizontal to a line of sight.
Angle of depression: The angle measured downwards from the horizontal to a line of sight.
Bearing: A direction measured clockwise from north, always written with three figures.
Perpendicular distance: The distance from a point to a line measured at right angles — the shortest such distance.

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