Ext Maths · 6 Trigonometry

6.2 Right-angled triangles

Before you start

Pythagoras' theorem and how to recognise the hypotenuse (6.1).
Rearranging a simple equation, including one where the unknown is in the denominator.
Using sin, cos, tan and their inverse keys on a scientific calculator set to degrees.
Bearings: measured clockwise from north and written with three figures.

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 6

Label the sides from the angle, every time

Trigonometry links an angle in a right-angled triangle to the ratio of two of its sides. Which two sides depends on where the angle is, so labelling comes first.
Mark the angle you are using (the given one, or the one you want). Then name the three sides from that angle's point of view.
Opposite faces the angle, adjacent touches it, and the hypotenuse faces the right angle.
Hypotenuse (H): opposite the right angle. It is the same side whichever acute angle you use.
Opposite (O): the side directly across from your angle. It does not touch your angle.
Adjacent (A): the side that touches both your angle and the right angle.
If you switch to the other acute angle, opposite and adjacent swap. That is why you label for one angle, not for the triangle.
The three ratios are not on the List of formulas, so learn them. SOH CAH TOA is the usual way to remember them.
$\sin θ = \frac{O}{H} \cos θ = \frac{A}{H} \tan θ = \frac{O}{A}$
To choose one: circle the side you know and the side you want (or the two sides you know, if you want the angle). Pick the ratio that uses exactly those two letters.
These ratios are for right-angled triangles only. Triangles without a right angle need the sine and cosine rules, which come in 6.5.

Part 2 of 6

Finding a side

Write the ratio with the numbers in, then rearrange. Where the unknown sits in the fraction decides whether you multiply or divide.
Unknown on top. $\sin 42 ° = \frac{x}{15}$ . The $x$ is being divided by $15$ , so multiply both sides by $15$ : $x = 15 \sin 42 °$ .
Unknown on the bottom. $\cos 40 ° = \frac{8}{x}$ . Multiply both sides by $x$ to get $x \cos 40 ° = 8$ , then divide by $\cos 40 °$ :
$x = \frac{8}{\cos 40 °}$
The trap is writing $x = 8 \cos 40 °$ for that second one. A quick check catches it: if $x$ is the hypotenuse, it must come out longer than $8$ .
Type the whole calculation in one go, such as $8 \div \cos 40$ . Rounding $\cos 40 °$ to a couple of decimals first can cost the accuracy mark.
Lengths go to 3 significant figures unless the question asks for something else.

Part 3 of 6

Finding an angle

If you know two sides, write the ratio as a fraction, then undo sin, cos or tan with the inverse key: $\sin^{- 1}$ , $\cos^{- 1}$ or $\tan^{- 1}$ (usually SHIFT then sin, cos or tan).
For example, adjacent $7$ and hypotenuse $11$ gives $\cos θ = \frac{7}{11}$ , so $θ = \cos^{- 1} (\frac{7}{11}) = 50.5 °$ .
Put the fraction straight into the inverse key, as in $\cos^{- 1} (7 \div 11)$ . Don't round the fraction to $0.64$ first.
The syllabus is explicit: angles are answered in degrees, correct to one decimal place. So $50.5 °$ , not $50 °$ and not $50.48 °$ .
Check your calculator shows D (degrees). In radian or gradian mode the answers are wrong but look believable.
Sense check: sine and cosine of an acute angle are always less than $1$ , because the hypotenuse is the longest side. If your fraction for sin or cos is more than $1$ , you have put the hypotenuse in the wrong place.

Part 4 of 6

Angles of elevation and depression

An angle of elevation is how far you look up, measured from the horizontal. For example, from the ground to the top of a tower.
An angle of depression is how far you look down, measured from the horizontal. For example, from a cliff top to a boat.
Both are measured from the horizontal, never from a vertical wall or cliff. Measure from the vertical and you get $90 °$ minus the angle you want.
The angle of depression from the top $T$ down to a point $B$ equals the angle of elevation from $B$ up to $T$ .
The two angles are alternate angles between parallel horizontals, so they are equal, and the one at B is inside the right-angled triangle.
Why they are equal: the horizontal through $T$ and the horizontal at sea level are parallel, and the line of sight crosses both. They are alternate angles.
That fact matters because the angle of depression sits outside the triangle, at the top. Its equal partner at the bottom sits inside the right-angled triangle, where SOH CAH TOA can use it.
Always sketch first: draw the horizontal, mark the angle against it, mark the right angle where the vertical meets the ground, then label O, A and H.

Part 5 of 6

Bearings and two-step problems

The syllabus notes say bearings may be required. A bearing is measured clockwise from north and written with three figures: $062 °$ , not $62 °$ .
To use trigonometry with a bearing, draw a north line at the start point, draw the journey, and drop a line east–west or north–south to make a right angle.
For a journey of length $d$ on a bearing $θ$ between $000 °$ and $090 °$ , the angle between the path and north is $θ$ itself. North is then adjacent and east is opposite.
$east = d \sin θ north = d \cos θ$
For any other bearing, don't memorise extra rules. Draw it and find the acute angle between the path and the north–south line. For a bearing of $125 °$ that angle is $180 ° - 125 ° = 55 °$ , measured from south.
The back bearing (from the finish back to the start) differs by $180 °$ . If $B$ is on a bearing of $062 °$ from $A$ , then $A$ is on a bearing of $242 °$ from $B$ .
Many problems need two steps: Pythagoras to find a length, then trigonometry for an angle, or two triangles that share a side.
Pythagoras uses three sides and no angle. Trigonometry is needed as soon as an angle is given or asked for.
Carry the unrounded value from the first step into the second. Round only at the very end.

Part 6 of 6

The shortest distance from a point to a line

The perpendicular distance from a point to a line is the shortest distance to the line.
Why: from a point $T$ , drop a perpendicular to the line, meeting it at $N$ . Any other route from $T$ to the line, say to $P$ , makes a right-angled triangle $T N P$ .
In that triangle $T P$ is the hypotenuse and $T N$ is a shorter side. The hypotenuse is always the longest side, so $T P$ is longer than $T N$ . This works for every choice of $P$ .
So when a question asks how close something gets to a road, a path or a ship's course, draw the perpendicular. That creates the right angle you need.
If $T P$ and the angle between $T P$ and the line are known, the perpendicular distance is the side opposite that angle: $T N = T P \sin (angle)$ .

Worked examples

Example 1. In a right-angled triangle, one angle is $34 °$ and the side opposite it is $5.8$ cm. Calculate the length of the hypotenuse.

Show the working

Label from the $34 °$ angle: $5.8$ cm is opposite, and you want the hypotenuse.
Opposite and hypotenuse means sine (SOH).
$\sin 34 ° = \frac{5.8}{h}$
The unknown is on the bottom. Multiply both sides by $h$ , then divide by $\sin 34 °$ .
$h = \frac{5.8}{\sin 34 °}$
Evaluate in one go.
$h = 10.372 \dots$
Round to 3 s.f.: $10.4$ cm. Check that the hypotenuse is longer than the $5.8$ cm side. It is. If you had multiplied instead, you would have got $3.24$ cm, which is shorter and so impossible.

Answer: $10.4$ cm (3 s.f.).

Example 2. A right-angled triangle has a hypotenuse of $17.3$ cm. The side adjacent to angle $θ$ is $12.5$ cm. Calculate $θ$ .

Show the working

The sides you know are the adjacent and the hypotenuse, so use cosine (CAH).
$\cos θ = \frac{12.5}{17.3}$
Undo cosine with the inverse key, putting the fraction straight in.
$θ = \cos^{- 1} (\frac{12.5}{17.3}) = 43.735 \dots °$
Angles go to one decimal place: $43.7 °$ .
Check: $\frac{12.5}{17.3}$ is less than $1$ , as a cosine must be, and the angle is acute.

Answer: $θ = 43.7 °$ .

Example 3. From the top of a vertical cliff $45$ m high, the angle of depression of a boat is $18 °$ . The boat sails directly towards the foot of the cliff until the angle of depression is $31 °$ . Calculate how far the boat sails.

Show the working

Sketch the cliff, sea level and the two boat positions. The angle of depression at the top equals the angle of elevation at the boat (alternate angles), so each triangle has its angle at the boat.
First position: the $45$ m cliff is opposite the $18 °$ angle, and the distance $d_{1}$ from the foot is adjacent. Opposite and adjacent means tangent.
$\tan 18 ° = \frac{45}{d_{1}} \Rightarrow d_{1} = \frac{45}{\tan 18 °} = 138.495 \dots m$
Second position, same idea with $31 °$ .
$d_{2} = \frac{45}{\tan 31 °} = 74.892 \dots m$
The boat sails the difference. Subtract the unrounded values.
$d_{1} - d_{2} = 138.495 \dots - 74.892 \dots = 63.603 \dots$
So the boat sails $63.6$ m. Watch what early rounding does: $138 - 74.9 = 63.1$ , which is wrong at 3 s.f.
Check: a bigger angle of depression means the boat is closer, so $d_{2} < d_{1}$ . It is.

Answer: $63.6$ m (3 s.f.).

Example 4. A walker starts at $A$ and ends $15$ km east and $8$ km north of $A$ . Calculate the direct distance from $A$ , and the bearing of the finish from $A$ .

Show the working

Sketch a north line at $A$ . The $8$ km north and $15$ km east steps meet at a right angle, and the direct path is the hypotenuse.
Distance by Pythagoras.
$d = \sqrt{15^{2} + 8^{2}} = \sqrt{289} = 17 km$
The bearing is the angle at $A$ between north and the path, measured clockwise. From that angle, the $15$ km east side is opposite and the $8$ km north side is adjacent.
$\tan θ = \frac{15}{8} \Rightarrow θ = \tan^{- 1} (\frac{15}{8}) = 61.927 \dots °$
Write the bearing with three figures before the decimal point: $061.9 °$ .
Check: more east than north means the bearing is between $045 °$ and $090 °$ . It is.

Answer: $17$ km on a bearing of $061.9 °$ .

Where people go wrong

Mistake: Labelling opposite and adjacent before deciding which angle you are working from.

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

Opposite and adjacent swap when you move to the other acute angle. Only the hypotenuse stays the same.

Mistake: Writing $x = 8 \cos 40 °$ from $\cos 40 ° = \frac{8}{x}$ .

Instead: When the unknown is on the bottom, divide: $x = \frac{8}{\cos 40 °}$ .

Multiplying both sides by $x$ and then dividing by $\cos 40 °$ is the only rearrangement that keeps the equation balanced. A hypotenuse shorter than $8$ is the warning sign.

Mistake: Measuring an angle of elevation or depression from the vertical.

Instead: Measure both from the horizontal.

From the vertical you get the complement, $90 °$ minus the correct angle, and every later step inherits the error.

Mistake: Rounding the first distance in a two-step problem before subtracting.

Instead: Subtract the full calculator values, then round.

In the cliff example, early rounding gives $63.1$ m instead of $63.6$ m, which loses the accuracy mark.

Mistake: Giving an angle to 3 significant figures or to the nearest degree.

Instead: Give angles correct to one decimal place.

The syllabus sets 1 d.p. for angles, separate from the 3 s.f. rule for lengths.

Check yourself

Write down the three trigonometric ratios using O, A and H.
Answer
$\sin θ = \frac{O}{H}$ , $\cos θ = \frac{A}{H}$ , $\tan θ = \frac{O}{A}$ .
Calculate $x$ when $\sin 42 ° = \frac{x}{15}$ .
Answer
$x = 15 \sin 42 ° = 10.0$ (3 s.f.).
A right-angled triangle has adjacent side $7$ cm and hypotenuse $11$ cm. Calculate the angle.
Answer
$\cos θ = \frac{7}{11}$ , so $θ = 50.5 °$ .
Explain why the angle of depression from a cliff top to a boat equals the angle of elevation from the boat to the cliff top.
Answer
Both are measured from horizontal lines, and the two horizontals are parallel. The line of sight crosses both, so the angles are alternate angles and therefore equal.
A tree $T$ is $60$ m from a point $P$ on a straight path. The angle between $P T$ and the path is $27 °$ . Calculate the shortest distance from the tree to the path.
Answer
The shortest distance is the perpendicular, which is opposite the $27 °$ angle with $P T$ as hypotenuse: $60 \sin 27 ° = 27.2$ m (3 s.f.).
Explain why the perpendicular distance from a point to a line is the shortest distance.
Answer
Any other line from the point to the line is the hypotenuse of a right-angled triangle that has the perpendicular as one of its shorter sides, and the hypotenuse is always the longest side.
The bearing of $B$ from $A$ is $062 °$ . Write down the bearing of $A$ from $B$ .
Answer
$062 ° + 180 ° = 242 °$ .

Exam questions

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

Question 1. In triangle $A B C$ , angle $A B C = 90 °$ , $A B = 7.4$ cm and $B C = 11.2$ cm.

(a) Calculate the length of $A C$ . 2 marks

Mark scheme

M1 7.4² + 11.2² oe, or 180.2 seen
A1 13.4 awrt (cm)

(b) Calculate angle $B A C$ . 2 marks

Mark scheme

M1 tan BAC = 11.2/7.4 oe, or a correct sin or cos using their AC
A1 56.5 awrt

(c) $D$ is the point on $A C$ such that $B D$ is perpendicular to $A C$ . Explain why $B D$ is the shortest distance from $B$ to the line $A C$ . 1 mark No calculator

Mark scheme

B1 any other line from B to AC is the hypotenuse of a right-angled triangle containing BD, so it is longer oe

(d) Calculate the length of $B D$ . 2 marks

Mark scheme

M1 7.4 sin(their BAC) oe, or 7.4 × 11.2 ÷ their AC
A1 6.17 awrt (cm); FT their (a) or (b)

Question 2. Maya stands on level ground, $32$ m from the foot of a vertical tower. The angle of elevation of the top of the tower from where she stands is $41 °$ . Ignore Maya's height.

(a) Calculate the height of the tower. 2 marks

Mark scheme

M1 32 tan 41° oe
A1 27.8 awrt (m)

(b) Maya walks directly away from the tower until the angle of elevation of the top is $25 °$ . Calculate how far she walks. 3 marks

Mark scheme

M1 their (a) ÷ tan 25° oe, or 59.7 awrt seen
M1 their distance − 32; dep on first M1
A1 27.7 awrt (m); 27.6 from premature rounding scores M1 M1 A0

(c) Write down the angle of depression of Maya's first position from the top of the tower. Give a reason for your answer. 2 marks No calculator

Mark scheme

B1 41°
B1 alternate angles, between parallel horizontal lines oe

Question 3. A ship leaves port $P$ and sails $18$ km on a bearing of $125 °$ to a point $Q$ .

(a) Calculate how far east of $P$ the point $Q$ is. 2 marks

Mark scheme

M1 18 sin 55° or 18 cos 35° oe
A1 14.7 awrt (km)

(b) Calculate how far south of $P$ the point $Q$ is. 2 marks

Mark scheme

M1 18 cos 55° or 18 sin 35° oe, or Pythagoras with their (a)
A1 10.3 awrt (km)

(c) Write down the bearing of $P$ from $Q$ . 1 mark No calculator

Mark scheme

B1 305° cao

How this is examined

Two triangles sharing a height: moving closer or further away, or two boats seen from one cliff: A 3- to 5-mark Calculate about a tower, cliff or building, where the height is found in one triangle and then used in a second. A correct trig statement for each triangle, such as $h = 32 \tan 41 °$ and then $\frac{h}{\tan 25 °}$ . Then subtracting to get the distance moved, using unrounded values. Early rounding usually loses only the final A mark, but it does lose it.
Choosing the ratio and rearranging correctly when the unknown is the hypotenuse: A 2- or 3-mark Calculate for a length inside a longer question. The method mark is for the correct rearranged form with numbers in, such as $\frac{5.8}{\sin 34 °}$ . Writing $5.8 \sin 34 °$ does not earn it, even though it uses the right ratio.
Giving angles to one decimal place: Every Calculate that ends in an angle. The accuracy mark needs an answer that rounds to the right value at 1 d.p., for example $56.5 °$ . Writing $57 °$ alone does not show that accuracy.
Turning a bearing into a right-angled triangle: A ship or walker question asking how far east, north or south something is, followed by a Write down for a back bearing. Finding the acute angle to the north–south line, such as $180 ° - 125 ° = 55 °$ , then using sin or cos correctly. A back bearing needs all three figures, such as $305 °$ .
Recognising 'shortest distance' as a perpendicular: A part asking how close a point gets to a line, sometimes with a 1-mark Explain. Drawing or naming the perpendicular, then using it as a side of a right-angled triangle. The Explain mark needs the idea that any other route is a hypotenuse, which is longer.

Words to know

Opposite: The side across from the angle you are using; it does not touch that angle.
Adjacent: The side touching both the angle you are using and the right angle.
Inverse function: $\sin^{- 1}$ , $\cos^{- 1}$ or $\tan^{- 1}$ : turns a ratio back into an 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, written with three figures.
Perpendicular distance: The distance from a point to a line measured at right angles to the line; the shortest distance.

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