Open Math›Year 9›Trigonometry›Angles of elevation and depression

Angles of elevation and depression

Identify angles of elevation and depression, draw and label right-angled triangles from real-world scenarios, and use trigonometric ratios to find unknown distances and heights.

Worked examples

Identifying angles of elevation and depression

Straightforward

Problem

A person stands

15

m from the base of a tree and looks up at the top at an angle of

40°

above the horizontal. (a) Name the type of angle formed. (b) Identify which side of the right-angled triangle represents the height of the tree, relative to the

40°

angle.

Determine whether the person is looking up or down from horizontal.

The person is looking upward from the horizontal, so the angle is an angle of elevation.

Sketch the right-angled triangle. The

40°

angle sits at the observer's position (on the ground). Label the three sides.

\text{hypotenuse} = \text{line of sight}, \quad \text{adjacent} = 15 \text{ m (horizontal base)}, \quad \text{opposite} = h \text{ (height of tree)}

Identify the height of the tree relative to the

40°

angle.

The height of the tree is directly across from the

40°

angle, so it is the opposite side.

Set up the ratio to find the height (for reference).

\tan 40° = \frac{h}{15} \implies h = 15 \times \tan 40° = 15 \times 0.8391\ldots \approx 12.59 \text{ m}

Answer

The angle is an angle of elevation. The height of the tree is the side opposite the

40°

angle.

Finding a height using angle of elevation

Moderate

Problem

From a point

40

m from the base of a vertical tower, the angle of elevation to the top is

38°

. Find the height of the tower, correct to 2 decimal places.

Identify the known information and the unknown.

\text{adjacent} = 40 \text{ m}, \quad \theta = 38°, \quad \text{opposite} = h

Choose the trigonometric ratio. The adjacent side (

40

m) and opposite side (

h

) are involved, so use TOA.

\tan 38° = \frac{h}{40}

Rearrange to make

h

the subject.

h = 40 \times \tan 38°

Evaluate using a calculator and round to 2 decimal places.

h = 40 \times 0.78129\ldots = 31.2514\ldots \approx 31.25 \text{ m}

Answer

h \approx 31.25

Multi-step: distance between two boats from a cliff

Challenging

Problem

From the top of a cliff

60

m above sea level, the angles of depression to two boats directly out to sea are

25°

(farther boat) and

40°

(nearer boat). Find the distance between the two boats, correct to 2 decimal places.

Set up the diagram. The cliff is

60

m high. Two lines of sight go to the two boats. Using alternate interior angles, the angle at the base of each right-angled triangle equals the angle of depression.

\text{For each boat: } \tan(\text{angle}) = \frac{60}{d} \implies d = \frac{60}{\tan(\text{angle})}

Find the horizontal distance from the cliff base to the nearer boat.

d_1 = \frac{60}{\tan 40°} = \frac{60}{0.83910\ldots} = 71.51 \text{ m}

Find the horizontal distance from the cliff base to the farther boat.

d_2 = \frac{60}{\tan 25°} = \frac{60}{0.46631\ldots} = 128.67 \text{ m}

Subtract to find the distance between the two boats.

\text{distance} = d_2 - d_1 = 128.67 - 71.51 = 57.16 \text{ m}

Answer

The distance between the two boats is approximately

57.16

Practise

Q1·Straightforward

A hiker stands on flat ground and looks up at the top of a mountain. The angle between the horizontal and the hiker's line of sight is called the angle of...

elevationdepressioninclinationrefraction

Explanation

When looking upward from horizontal, the angle formed is an angle of elevation. The angle of depression is formed when looking downward from horizontal.

Q2·Straightforward

A pilot looks down from an aircraft at a runway far below. The angle between the horizontal and the pilot's line of sight to the runway is called the angle of...

elevationdepressiondeclinationinclination

Explanation

When looking downward from horizontal, the angle formed is an angle of depression. It is measured between the horizontal and the line of sight going down.

Q3·Straightforward

A person stands

20

m from the base of a tower and looks up at the top at an angle of elevation of

45°

. In the right-angled triangle formed, the height of the tower is the side...

opposite the $45°$ angleadjacent to the $45°$ anglethe hypotenuseequal to the $20$ m base

Explanation

The

45°

angle is at the observer's position. The height of the tower is the vertical side directly across from this angle, so it is the opposite side. The

20

m base is the adjacent side and the line of sight is the hypotenuse.

Q4·Straightforward

From a window

12

m above the ground, a person looks down at a point on the ground at an angle of depression of

55°

. Which trigonometric ratio should be used to find the horizontal distance to that point?

$\sin$$\cos$$\tan$$\cos^{-1}$

Explanation

The height (

12

m) is opposite the

55°

angle and the horizontal distance

d

is adjacent. Using TOA:

\tan 55° = \dfrac{12}{d}

, so

\tan

is the correct ratio.

Q5·Moderate

A person stands

30

m from the base of a building and measures the angle of elevation to the top as

42°

. Find the height of the building, correct to 2 decimal places.

Explanation

Using TOA:

\tan 42° = \dfrac{h}{30}

, so

h = 30 \times \tan 42° = 30 \times 0.9004\ldots = 27.01

Q6·Moderate

From the top of a cliff

75

m high, the angle of depression to a boat at sea is

18°

. Find the horizontal distance from the base of the cliff to the boat, correct to 2 decimal places.

Explanation

Using alternate interior angles, the

18°

angle also appears at the base of the right-angled triangle. The cliff (

75

m) is opposite the

18°

angle and the horizontal distance

d

is adjacent. Using TOA:

\tan 18° = \dfrac{75}{d}

, so

d = \dfrac{75}{\tan 18°} = \dfrac{75}{0.3249\ldots} = 230.83

Q7·Moderate

A ladder

5.5

m long leans against a vertical wall, making an angle of elevation of

68°

with the ground. Find how high up the wall the ladder reaches, correct to 2 decimal places.

Explanation

Using SOH:

\sin 68° = \dfrac{h}{5.5}

, so

h = 5.5 \times \sin 68° = 5.5 \times 0.9272\ldots = 5.10

Q8·Moderate

A kite is flying on a

50

m string that makes an angle of elevation of

62°

with the ground. Assuming the string is straight, find the height of the kite above the ground, correct to 2 decimal places.

Explanation

Using SOH:

\sin 62° = \dfrac{h}{50}

, so

h = 50 \times \sin 62° = 50 \times 0.8829\ldots = 44.15

Q9·Challenging

From the top of a cliff

50

m above sea level, the angles of depression to two boats directly out to sea are

20°

(farther boat) and

35°

(nearer boat). Find the distance between the two boats, correct to 2 decimal places.

Explanation

Horizontal distance to nearer boat:

d_1 = \dfrac{50}{\tan 35°} = \dfrac{50}{0.7002\ldots} \approx 71.41

m. Horizontal distance to farther boat:

d_2 = \dfrac{50}{\tan 20°} = \dfrac{50}{0.3640\ldots} \approx 137.37

m. Distance between boats:

137.37 - 71.41 \approx 65.97

Q10·Challenging

From a point P on the ground, the angle of elevation to the top of a tower is

32°

. From a point Q,

30

m closer to the tower along the same horizontal line, the angle of elevation is

48°

. Find the height of the tower, correct to 2 decimal places.

Explanation

Let

h

= height and

d

= horizontal distance from Q to tower base. Then

\tan 48° = \dfrac{h}{d}

and

\tan 32° = \dfrac{h}{d + 30}

. From the first equation,

d = \dfrac{h}{\tan 48°}

. Substituting:

\tan 32° = \dfrac{h}{\dfrac{h}{\tan 48°} + 30}

. Rearranging gives

h = \dfrac{30}{\cot 32° - \cot 48°} = \dfrac{30}{1.6003 - 0.9004} = \dfrac{30}{0.6999} \approx 42.86

Q11·Challenging

From the top of a

45

m building, the angle of depression to the base of a shorter building directly across a street is

62°

. The angle of depression from the top of the taller building to the roof of the shorter building is

48°

. Find the height of the shorter building, correct to 2 decimal places.

Explanation

Street width:

d = \dfrac{45}{\tan 62°} = \dfrac{45}{1.8807\ldots} \approx 23.93

m. The vertical drop from the top of the taller building to the roof of the shorter:

23.93 \times \tan 48° = 23.93 \times 1.1106\ldots \approx 26.57

m. Height of shorter building:

45 - 26.57 \approx 18.43

Q12·Challenging

A flagpole sits on top of a

10

m wall. From a point on flat ground

20

m from the base of the wall, the angle of elevation to the top of the flagpole is

55°

. Find the height of the flagpole, correct to 2 decimal places.

Explanation

Let

p

= height of the flagpole. The total height from the ground to the top of the flagpole is

10 + p

. Using TOA:

\tan 55° = \dfrac{10 + p}{20}

, so

10 + p = 20 \times \tan 55° = 20 \times 1.4281\ldots = 28.56

. Therefore

p = 28.56 - 10 = 18.56