Trigonometry, Sine & Cosine Rule, Bearings: O-Level & SEC E-Math

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Maths In Real Life

Real situations, each with its own picture. The pictures are drawn first, then the maths is placed on top.

$A gentle ramp beside two steps at the open ground floor of an apartment block, with the horizontal run, the vertical rise and the ramp length drawn over it$

Maths In Real Life 1

A ramp at the void deck

A ramp lets prams and wheelchairs roll up a small height without steps. The gentler the angle, the easier the climb, but the longer the ramp has to be.

A ramp rises 0.45 m over a horizontal distance of 5.4 m.

(a)Find the length of the ramp, to three significant figures.
(b)Find the angle the ramp makes with the ground, to one decimal place.

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(a)

\displaystyle \text{length}^2 = 5.4^2 + 0.45^2 = 29.3625

Pythagoras.

\displaystyle \text{length} = \sqrt{29.3625} \approx 5.4187

Square root.

\displaystyle \text{length} \approx 5.42

Round to the accuracy the question asks for.

(b)

\displaystyle \tan\theta = \dfrac{0.45}{5.4} \approx 0.083333

Opposite over adjacent.

\displaystyle \theta = \tan^{-1}(0.083333) \approx 4.7636^\circ

Inverse tangent.

\displaystyle \theta \approx 4.8^\circ

Round to the accuracy the question asks for.

Final answer

(a)The ramp is 5.42 m long.

(b)The angle is 4.8 degrees.

$A teenager measuring the height of an apartment block with a phone, with the horizontal distance, eye height and angle of elevation drawn over the picture$

Maths In Real Life 2

Measuring a block with a phone

Free clinometer apps use the phone's tilt sensor to show an angle of elevation. Combined with a measured distance, that is enough to find the height of a building.

A student stands 60 m from the foot of a block of flats on level ground. Her phone reads an angle of elevation of 49 degrees to the top of the block. Her eye is 1.55 m above the ground.

Find the height of the block, correct to three significant figures.

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\displaystyle \tan 49^\circ = \dfrac{h}{60}

h

is the height above her eye level.

\displaystyle h = 60 \times 1.1504 \approx 69.022

Multiply.

\displaystyle 69.022 + 1.55 = 70.572

Add her eye height.

\displaystyle 69.022 + 1.55 \approx 70.6

Round to the accuracy the question asks for.

Final answer

The block is 70.6 m tall.

$A patrol boat and a fishing boat leaving a harbour in different directions, with a North line, the two bearings and the distance between the boats drawn over the picture$

Maths In Real Life 3

Two boats leaving harbour

Coast guards track boats by their bearing and distance from a base. Two bearings and two distances are enough to work out how far apart the boats are.

A patrol boat leaves a harbour H and travels 12 km on a bearing of 035°. A fishing boat leaves H and travels 9 km on a bearing of 112°.

Find the distance between the boats, correct to three significant figures.

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\displaystyle \angle PHF = 112^{\circ} - 035^{\circ} = 77^\circ

The angle between the two bearings.

\displaystyle d^2 = 12^2 + 9^2 - 2 \times 12 \times 9 \times \cos 77^\circ

Cosine rule.

\displaystyle d^2 = 12^2 + 9^2 - 2 \times 12 \times 9 \times 0.22495 \approx 176.41

Evaluate.

\displaystyle d = \sqrt{176.41} \approx 13.282

Square root.

\displaystyle d \approx 13.3

Round to the accuracy the question asks for.

Final answer

The boats are 13.3 km apart.

Supports G4.1 · G4.2

Pythagoras' theorem and its converse

Quick recap. In a right-angled triangle the side opposite the right angle is the hypotenuse, and it is always the longest side. Pythagoras' theorem links the three sides: the square on the hypotenuse equals the sum of the squares on the other two sides.

The Core Idea

Square the two short sides and add to get the long side squared.

If the legs are $a$ and $b$ and the hypotenuse is $c$ , then $c^2 = a^2 + b^2$ . To find a leg you subtract: $a^2 = c^2 - b^2$ . Before you press the calculator, ask: is the unknown side the longest side or not?

SGSchoolKaki Tip

Pythagoras needs a right angle. If the question gives you no right angle and no way to prove one, it is a job for the sine rule or the cosine rule (later in this chapter).

Think About It

Why can the hypotenuse never be shorter than a leg?

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Pythagoras says the square on the hypotenuse is the sum of the squares on the two legs.

Start from the theorem.

The square on the other leg is a positive amount, so the sum is bigger than the square on this leg.

Something positive is added.

A bigger square means a bigger side, so the hypotenuse is longer than each leg.

Compare the sides themselves.

Final answerA right-angled triangle always has its longest side opposite the right angle, so no leg can beat it.

Example 13.1

Finding the hypotenuse

A straight ramp rises 6 m over a horizontal distance of 9 m. Find the length of the ramp, correct to three significant figures.

Find the length of the ramp.

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Sketch the right-angled triangle. The ramp is opposite the right angle, so it is the hypotenuse

c

.

Spot the hypotenuse first.

\displaystyle c^2 = a^2 + b^2

Pythagoras' theorem.

\displaystyle c^2 = 9^2 + 6^2

Substitute the two shorter sides.

\displaystyle c^2 = 117

Add the squares.

\displaystyle c = \sqrt{117} \approx 10.817

Take the square root; keep extra figures in the working.

\displaystyle c \approx 10.8

Round to the accuracy the question asks for.

Final answer

The ramp is 10.8 m long.

Your Turn 13.1

A straight ramp rises 3.5 m over a horizontal distance of 8.4 m. Find the length of the ramp, correct to three significant figures.

Find the length of the ramp.

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Sketch the right-angled triangle. The ramp is opposite the right angle, so it is the hypotenuse

c

.

Spot the hypotenuse first.

\displaystyle c^2 = a^2 + b^2

Pythagoras' theorem.

\displaystyle c^2 = 8.4^2 + 3.5^2

Substitute the two shorter sides.

\displaystyle c^2 = 82.81

Add the squares.

\displaystyle c = \sqrt{82.81} = 9.1

Take the square root; keep extra figures in the working.

\displaystyle c \approx 9.10

Round to the accuracy the question asks for.

Final answer

The ramp is 9.10 m long.

Example 13.2

Finding a shorter side

A ladder of length 7.2 m rests against a vertical wall. Its foot is 2.1 m from the wall on level ground. How high up the wall does the ladder reach? Give your answer to three significant figures.

Find the height reached.

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The ladder is the hypotenuse, so the height is a shorter side. Subtract the squares.

Unknown is a leg, so subtract.

\displaystyle h^2 = c^2 - a^2

Rearrange Pythagoras' theorem.

\displaystyle h^2 = 7.2^2 - 2.1^2 = 51.84 - 4.41

Square each given side.

\displaystyle h^2 = 47.43

Subtract.

\displaystyle h = \sqrt{47.43} \approx 6.8869

Square root.

\displaystyle h \approx 6.89

Round to the accuracy the question asks for.

Final answer

The ladder reaches 6.89 m up the wall.

Your Turn 13.2

A ladder of length 6.5 m rests against a vertical wall. Its foot is 1.8 m from the wall on level ground. How high up the wall does the ladder reach? Give your answer to three significant figures.

Find the height reached.

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The ladder is the hypotenuse, so the height is a shorter side. Subtract the squares.

Unknown is a leg, so subtract.

\displaystyle h^2 = c^2 - a^2

Rearrange Pythagoras' theorem.

\displaystyle h^2 = 6.5^2 - 1.8^2 = 42.25 - 3.24

Square each given side.

\displaystyle h^2 = 39.01

Subtract.

\displaystyle h = \sqrt{39.01} \approx 6.2458

Square root.

\displaystyle h \approx 6.25

Round to the accuracy the question asks for.

Final answer

The ladder reaches 6.25 m up the wall.

Example 13.3

Is the triangle right-angled?

Triangle P has sides 20 cm, 21 cm and 29 cm. Triangle Q has sides 9 cm, 12 cm and 16 cm. Decide, with working, which triangle is right-angled.

(i)Is triangle P right-angled?
(ii)Is triangle Q right-angled?

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(i)

Test the converse: if the square of the longest side equals the sum of the squares of the other two, the triangle has a right angle opposite it.

The longest side is the only possible hypotenuse.

\displaystyle 20^2 + 21^2 = 841

Add the squares of the two shorter sides.

\displaystyle 29^2 = 841

Square the longest side.

The two results are equal, so triangle P is right-angled.

Equal means the converse holds.

(ii)

\displaystyle 9^2 + 12^2 = 225

Short sides first.

\displaystyle 16^2 = 256

Longest side squared.

The results are different, so triangle Q is not right-angled.

One mismatch is enough to say no.

Final answer

(i)Triangle P is right-angled: 841 = 841.

(ii)Triangle Q is not right-angled: 225 is not 256.

Your Turn 13.3

Triangle P has sides 28 cm, 45 cm and 53 cm. Triangle Q has sides 10 cm, 11 cm and 15 cm. Decide, with working, which triangle is right-angled.

(i)Is triangle P right-angled?
(ii)Is triangle Q right-angled?

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(i)

Test the converse: if the square of the longest side equals the sum of the squares of the other two, the triangle has a right angle opposite it.

The longest side is the only possible hypotenuse.

\displaystyle 28^2 + 45^2 = 2809

Add the squares of the two shorter sides.

\displaystyle 53^2 = 2809

Square the longest side.

The two results are equal, so triangle P is right-angled.

Equal means the converse holds.

(ii)

\displaystyle 10^2 + 11^2 = 221

Short sides first.

\displaystyle 15^2 = 225

Longest side squared.

The results are different, so triangle Q is not right-angled.

One mismatch is enough to say no.

Final answer

(i)Triangle P is right-angled: 2809 = 2809.

(ii)Triangle Q is not right-angled: 221 is not 225.

Example 13.4

Pythagoras in an isosceles triangle

An isosceles triangle has two equal sides of 8.5 cm and a base of 7.2 cm. Find its perpendicular height and its area, each to three significant figures.

(i)Find the perpendicular height.
(ii)Find the area.

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(i)

The height drops from the apex to the midpoint of the base, making two right-angled triangles.

An isosceles triangle splits into two equal right-angled triangles.

\displaystyle \text{half the base} = 7.2 \div 2 = 3.6

The base is shared equally.

\displaystyle h^2 = 8.5^2 - 3.6^2 = 59.29

The equal side is the hypotenuse.

\displaystyle h = \sqrt{59.29} = 7.7

Square root.

\displaystyle h \approx 7.70

Round to the accuracy the question asks for.

(ii)

\displaystyle \text{Area} = \tfrac{1}{2} \times 7.2 \times 7.7 = 27.72

Half base times the unrounded height.

\displaystyle \text{Area} \approx 27.7

Round to the accuracy the question asks for.

Final answer

(i)The height is 7.70 cm.

(ii)The area is 27.7 square centimetres.

Your Turn 13.4

An isosceles triangle has two equal sides of 6.4 cm and a base of 5 cm. Find its perpendicular height and its area, each to three significant figures.

(i)Find the perpendicular height.
(ii)Find the area.

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(i)

The height drops from the apex to the midpoint of the base, making two right-angled triangles.

An isosceles triangle splits into two equal right-angled triangles.

\displaystyle \text{half the base} = 5 \div 2 = 2.5

The base is shared equally.

\displaystyle h^2 = 6.4^2 - 2.5^2 = 34.71

The equal side is the hypotenuse.

\displaystyle h = \sqrt{34.71} \approx 5.8915

Square root.

\displaystyle h \approx 5.89

Round to the accuracy the question asks for.

(ii)

\displaystyle \text{Area} = \tfrac{1}{2} \times 5 \times 5.8915 \approx 14.729

Half base times the unrounded height.

\displaystyle \text{Area} \approx 14.7

Round to the accuracy the question asks for.

Final answer

(i)The height is 5.89 cm.

(ii)The area is 14.7 square centimetres.

Links To

Pythagoras also drives the distance formula in Chapter 15, Coordinate geometry

Links To

Revise the basics for free in the free E-Math revision page: circles and trigonometry

Supports G4.3 · G4.1

Trigonometric ratios in right-angled triangles

Pick one acute angle in a right-angled triangle and call it $\theta$ . Relative to $\theta$ , the side across from it is the opposite, the side touching it (that is not the hypotenuse) is the adjacent, and the longest side is the hypotenuse.

The three sides, named from the angle

\theta

.

The Core Idea

SOH CAH TOA.

$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$ , $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ , $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$ . Write down which two sides are involved before choosing the ratio.

Case table: which ratio do I use?
Sides involved	Picture	Ratio	Use it when
opposite and hypotenuse	$\sin\theta$	sine: $\frac{\text{opp}}{\text{hyp}}$	you know one of these two and want the other, or the angle
adjacent and hypotenuse	$\cos\theta$	cosine: $\frac{\text{adj}}{\text{hyp}}$	you know one of these two and want the other, or the angle
opposite and adjacent	$\tan\theta$	tangent: $\frac{\text{opp}}{\text{adj}}$	the hypotenuse is not involved

SGSchoolKaki Tip

Set the calculator to degrees (D). To find an angle from a ratio use the inverse keys $\sin^{-1}$ , $\cos^{-1}$ , $\tan^{-1}$ . Keep the whole calculator value in the working and round only the final answer.

Example 13.5

Finding a side: unknown on top

In a right-angled triangle the hypotenuse is 12 cm and one acute angle is 35 degrees. Find the length $x$ of the side opposite that angle, correct to three significant figures.

Find $x$ .

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Label the sides from the given angle:

x

is opposite, and the hypotenuse is known. Opposite and hypotenuse means sine.

Name the two sides, then pick the ratio.

\displaystyle \sin 35^\circ = \dfrac{x}{12}

Write the ratio with the unknown on top.

\displaystyle x = 12 \times \sin 35^\circ

Multiply both sides by the hypotenuse.

\displaystyle x = 12 \times 0.57358

Calculator in degree mode.

\displaystyle x \approx 6.8829

Keep extra figures.

\displaystyle x \approx 6.88

Round to the accuracy the question asks for.

Final answer

x

= 6.88 cm.

Your Turn 13.5

In a right-angled triangle the hypotenuse is 9.5 cm and one acute angle is 52 degrees. Find the length $x$ of the side opposite that angle, correct to three significant figures.

Find $x$ .

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Label the sides from the given angle:

x

is opposite, and the hypotenuse is known. Opposite and hypotenuse means sine.

Name the two sides, then pick the ratio.

\displaystyle \sin 52^\circ = \dfrac{x}{9.5}

Write the ratio with the unknown on top.

\displaystyle x = 9.5 \times \sin 52^\circ

Multiply both sides by the hypotenuse.

\displaystyle x = 9.5 \times 0.78801

Calculator in degree mode.

\displaystyle x \approx 7.4861

Keep extra figures.

\displaystyle x \approx 7.49

Round to the accuracy the question asks for.

Final answer

x

= 7.49 cm.

Example 13.6

Finding a side: unknown on the bottom

A support wire is fixed to the top of a pole and makes an angle of 28 degrees with the level ground. The wire is anchored 9.4 m from the foot of the pole. Find the length $y$ of the wire, correct to three significant figures.

Find $y$ .

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Relative to the angle at the ground, the known side is the adjacent and

y

is the hypotenuse. Adjacent and hypotenuse means cosine.

Adjacent and hypotenuse: cosine.

\displaystyle \cos 28^\circ = \dfrac{9.4}{y}

The unknown is now on the bottom.

\displaystyle y = \dfrac{9.4}{\cos 28^\circ}

Swap

y

and

\cos

: multiply by

y

, then divide by the cosine.

\displaystyle y = \dfrac{9.4}{0.88295} \approx 10.646

Divide, do not multiply.

\displaystyle y \approx 10.6

Round to the accuracy the question asks for.

Final answer

The wire is 10.6 m long.

Your Turn 13.6

A support wire is fixed to the top of a pole and makes an angle of 41 degrees with the level ground. The wire is anchored 6.8 m from the foot of the pole. Find the length $y$ of the wire, correct to three significant figures.

Find $y$ .

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Relative to the angle at the ground, the known side is the adjacent and

y

is the hypotenuse. Adjacent and hypotenuse means cosine.

Adjacent and hypotenuse: cosine.

\displaystyle \cos 41^\circ = \dfrac{6.8}{y}

The unknown is now on the bottom.

\displaystyle y = \dfrac{6.8}{\cos 41^\circ}

Swap

y

and

\cos

: multiply by

y

, then divide by the cosine.

\displaystyle y = \dfrac{6.8}{0.75471} \approx 9.0101

Divide, do not multiply.

\displaystyle y \approx 9.01

Round to the accuracy the question asks for.

Final answer

The wire is 9.01 m long.

Example 13.7

Finding an angle

In a right-angled triangle the side opposite the angle $\theta$ is 7 cm and the hypotenuse is 11 cm. Find $\theta$ and the other acute angle, each correct to one decimal place.

(i)Find $\theta$ .
(ii)Find the other acute angle.

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(i)

Opposite and hypotenuse, so use sine.

Name the sides first.

\displaystyle \sin\theta = \dfrac{7}{11} \approx 0.63636

Write the ratio as a number.

\displaystyle \theta = \sin^{-1}(0.63636) \approx 39.521^\circ

Use the inverse sine key.

\displaystyle \theta \approx 39.5^\circ

Round to the accuracy the question asks for.

(ii)

\displaystyle 90^\circ - 39.521^\circ \approx 50.479^\circ

The two acute angles of a right-angled triangle add up to ninety degrees.

\displaystyle 90^\circ - 39.521^\circ \approx 50.5^\circ

Round to the accuracy the question asks for.

Final answer

(i)

\theta

= 39.5 degrees.

(ii)The other acute angle is 50.5 degrees.

Your Turn 13.7

In a right-angled triangle the side opposite the angle $\theta$ is 4.5 cm and the hypotenuse is 8 cm. Find $\theta$ and the other acute angle, each correct to one decimal place.

(i)Find $\theta$ .
(ii)Find the other acute angle.

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(i)

Opposite and hypotenuse, so use sine.

Name the sides first.

\displaystyle \sin\theta = \dfrac{4.5}{8} = 0.5625

Write the ratio as a number.

\displaystyle \theta = \sin^{-1}(0.5625) \approx 34.229^\circ

Use the inverse sine key.

\displaystyle \theta \approx 34.2^\circ

Round to the accuracy the question asks for.

(ii)

\displaystyle 90^\circ - 34.229^\circ \approx 55.771^\circ

The two acute angles of a right-angled triangle add up to ninety degrees.

\displaystyle 90^\circ - 34.229^\circ \approx 55.8^\circ

Round to the accuracy the question asks for.

Final answer

(i)

\theta

= 34.2 degrees.

(ii)The other acute angle is 55.8 degrees.

Example 13.8

Solving a whole right-angled triangle

In triangle PQR, angle PQR is a right angle, PQ = 7.5 cm and QR = 4.2 cm. Find angle QPR, PR and angle PRQ, to the usual accuracy (angles to one decimal place, lengths to three significant figures).

(i)Find angle QPR.
(ii)Find PR.
(iii)Find angle PRQ.

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(i)

Relative to angle QPR, QR is opposite and PQ is adjacent: use tangent.

Opposite and adjacent: tangent.

\displaystyle \tan P = \dfrac{4.2}{7.5} = 0.56

Ratio as a number.

\displaystyle P = \tan^{-1}(0.56) \approx 29.249^\circ

Inverse tangent.

\displaystyle P \approx 29.2^\circ

Round to the accuracy the question asks for.

(ii)

\displaystyle PR^2 = 7.5^2 + 4.2^2 = 73.89

Use Pythagoras, so the rounded angle is not carried into the length.

\displaystyle PR = \sqrt{73.89} \approx 8.5959

Square root.

\displaystyle PR \approx 8.60

Round to the accuracy the question asks for.

(iii)

\displaystyle 90^\circ - 29.249^\circ \approx 60.751^\circ

The angles of a triangle add up to

180^\circ

, and the right angle uses ninety of them.

\displaystyle 90^\circ - 29.249^\circ \approx 60.8^\circ

Round to the accuracy the question asks for.

Final answer

(i)Angle QPR = 29.2 degrees.

(ii)PR = 8.60 cm.

(iii)Angle PRQ = 60.8 degrees.

Your Turn 13.8

In triangle PQR, angle PQR is a right angle, PQ = 6.3 cm and QR = 5.9 cm. Find angle QPR, PR and angle PRQ, to the usual accuracy (angles to one decimal place, lengths to three significant figures).

(i)Find angle QPR.
(ii)Find PR.
(iii)Find angle PRQ.

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(i)

Relative to angle QPR, QR is opposite and PQ is adjacent: use tangent.

Opposite and adjacent: tangent.

\displaystyle \tan P = \dfrac{5.9}{6.3} \approx 0.93651

Ratio as a number.

\displaystyle P = \tan^{-1}(0.93651) \approx 43.122^\circ

Inverse tangent.

\displaystyle P \approx 43.1^\circ

Round to the accuracy the question asks for.

(ii)

\displaystyle PR^2 = 6.3^2 + 5.9^2 = 74.5

Use Pythagoras, so the rounded angle is not carried into the length.

\displaystyle PR = \sqrt{74.5} \approx 8.6313

Square root.

\displaystyle PR \approx 8.63

Round to the accuracy the question asks for.

(iii)

\displaystyle 90^\circ - 43.122^\circ \approx 46.878^\circ

The angles of a triangle add up to

180^\circ

, and the right angle uses ninety of them.

\displaystyle 90^\circ - 43.122^\circ \approx 46.9^\circ

Round to the accuracy the question asks for.

Final answer

(i)Angle QPR = 43.1 degrees.

(ii)PR = 8.63 cm.

(iii)Angle PRQ = 46.9 degrees.

Supports G4.7

Angles of elevation and depression

An angle of elevation is measured upwards from the horizontal to your line of sight. An angle of depression is measured downwards from the horizontal. Both are measured from a horizontal line, never from the vertical.

The Core Idea

Draw the horizontal first.

Mark the horizontal line through the observer, then the line of sight, then the angle between them. The right-angled triangle you need is usually hiding between the ground, the object and the line of sight.

SGSchoolKaki Tip

Do not measure the angle of elevation from the vertical wall or pole. It is always measured upwards from the horizontal ground.

Example 13.9

Angle of elevation

Ayesha stands 45 m from the foot of a tower on level ground. Her eyes are 1.6 m above the ground. She measures the angle of elevation of the top of the tower as 32 degrees. Find the height of the tower, correct to three significant figures.

Find the height of the tower.

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The horizontal through the eyes meets the tower. Above that point the tower forms a right-angled triangle with the line of sight. The side above eye level is opposite the angle, and the distance 45 m is adjacent.

Opposite and adjacent: tangent.

\displaystyle \tan 32^\circ = \dfrac{h}{45}

Let

h

be the height of the tower above eye level.

\displaystyle h = 45 \times 0.62487 \approx 28.119

Multiply by the adjacent side.

\displaystyle \text{Tower height} = 28.119 + 1.6 = 29.719

Add the eye height, because

h

stops at eye level.

\displaystyle \text{Tower height} \approx 29.7

Round to the accuracy the question asks for.

Final answer

The tower is 29.7 m tall.

Your Turn 13.9

Ayesha stands 62 m from the foot of a tower on level ground. Her eyes are 1.5 m above the ground. She measures the angle of elevation of the top of the tower as 27 degrees. Find the height of the tower, correct to three significant figures.

Find the height of the tower.

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The horizontal through the eyes meets the tower. Above that point the tower forms a right-angled triangle with the line of sight. The side above eye level is opposite the angle, and the distance 62 m is adjacent.

Opposite and adjacent: tangent.

\displaystyle \tan 27^\circ = \dfrac{h}{62}

Let

h

be the height of the tower above eye level.

\displaystyle h = 62 \times 0.50953 \approx 31.591

Multiply by the adjacent side.

\displaystyle \text{Tower height} = 31.591 + 1.5 = 33.091

Add the eye height, because

h

stops at eye level.

\displaystyle \text{Tower height} \approx 33.1

Round to the accuracy the question asks for.

Final answer

The tower is 33.1 m tall.

Example 13.10

Angle of depression

From a window 60 m above level ground, the angle of depression of a bicycle on the ground is 25 degrees. Find the horizontal distance from the foot of the block to the bicycle, correct to three significant figures.

Find the horizontal distance $d$ .

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The angle of depression is measured down from the horizontal. The horizontal line at the window is parallel to the ground, so the angle of elevation of the window from the bicycle is equal to it (alternate angles).

Alternate angles between parallel lines are equal.

\displaystyle \tan 25^\circ = \dfrac{60}{d}

Height is opposite,

d

is adjacent.

\displaystyle d = \dfrac{60}{\tan 25^\circ} = \dfrac{60}{0.46631} \approx 128.67

Unknown on the bottom: divide.

\displaystyle d \approx 129

Round to the accuracy the question asks for.

Final answer

The bicycle is 129 m from the foot of the block.

Your Turn 13.10

From a window 48 m above level ground, the angle of depression of a bicycle on the ground is 31 degrees. Find the horizontal distance from the foot of the block to the bicycle, correct to three significant figures.

Find the horizontal distance $d$ .

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The angle of depression is measured down from the horizontal. The horizontal line at the window is parallel to the ground, so the angle of elevation of the window from the bicycle is equal to it (alternate angles).

Alternate angles between parallel lines are equal.

\displaystyle \tan 31^\circ = \dfrac{48}{d}

Height is opposite,

d

is adjacent.

\displaystyle d = \dfrac{48}{\tan 31^\circ} = \dfrac{48}{0.60086} \approx 79.885

Unknown on the bottom: divide.

\displaystyle d \approx 79.9

Round to the accuracy the question asks for.

Final answer

The bicycle is 79.9 m from the foot of the block.

Example 13.11

Two observation points

Points A and B are on level ground in a straight line with the foot of a vertical tower, and B is between A and the tower. A and B are 50 m apart. The angle of elevation of the top of the tower is 20 degrees from A and 35 degrees from B. Find the height of the tower, correct to three significant figures.

Find the height $h$ .

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Let the distance from B to the foot of the tower be

t

. Both right-angled triangles share the height

h

.

Two triangles, two equations, one height.

\displaystyle h = t \tan 35^\circ \quad \text{and} \quad h = (t + 50)\tan 20^\circ

Tangent in each triangle (opposite over adjacent).

\displaystyle t = h \times \dfrac{1}{\tan 35^\circ} = h \times 1.4281

Make

t

the subject from the first equation.

\displaystyle t + 50 = h \times 2.7475

From the second equation.

\displaystyle h \times 2.7475 - h \times 1.4281 = 50

Subtract the two expressions for the horizontal distances.

\displaystyle h = \dfrac{50}{1.3193} \approx 37.898

Factor out

h

, then divide.

\displaystyle h \approx 37.9

Round to the accuracy the question asks for.

Final answer

The tower is 37.9 m tall.

Your Turn 13.11

Points A and B are on level ground in a straight line with the foot of a vertical tower, and B is between A and the tower. A and B are 60 m apart. The angle of elevation of the top of the tower is 25 degrees from A and 40 degrees from B. Find the height of the tower, correct to three significant figures.

Find the height $h$ .

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Colour keyadd, subtract, multiply, divide or simplifyround, estimate or compare

Let the distance from B to the foot of the tower be

t

. Both right-angled triangles share the height

h

.

Two triangles, two equations, one height.

\displaystyle h = t \tan 40^\circ \quad \text{and} \quad h = (t + 60)\tan 25^\circ

Tangent in each triangle (opposite over adjacent).

\displaystyle t = h \times \dfrac{1}{\tan 40^\circ} = h \times 1.1918

Make

t

the subject from the first equation.

\displaystyle t + 60 = h \times 2.1445

From the second equation.

\displaystyle h \times 2.1445 - h \times 1.1918 = 60

Subtract the two expressions for the horizontal distances.

\displaystyle h = \dfrac{60}{0.95275} \approx 62.975

Factor out

h

, then divide.

\displaystyle h \approx 63.0

Round to the accuracy the question asks for.

Final answer

The tower is 63.0 m tall.

Pythagoras' Theorem and Trigonometry: O-Level E-Math Notes

Maths In Real Life

A ramp at the void deck

Measuring a block with a phone

Two boats leaving harbour

Pythagoras' theorem and its converse

Finding the hypotenuse

Finding a shorter side

Is the triangle right-angled?

Pythagoras in an isosceles triangle

Trigonometric ratios in right-angled triangles

Finding a side: unknown on top

Finding a side: unknown on the bottom

Finding an angle

Solving a whole right-angled triangle

Angles of elevation and depression

Angle of elevation

Angle of depression

Two observation points

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