Heights and distances for CDS

Heights and distances is trigonometry with a picture. A tower, a cliff, a kite or an aeroplane stands in for the right triangle, and the question gives you one angle and one length. The maths is almost always a single tan, sin or cos from the standard table. The difficulty lies in drawing the figure correctly and choosing the right triangle.

In the CDS paper these questions usually come in the trigonometry block. The angles are nearly always 30°, 45° and 60°, so answers come out as whole numbers or multiples of √3. If your working gives something like tan 37°, you have probably set up the figure wrongly.

Elevation and depression

• Angle of elevation: you look up at an object. It is the angle between the horizontal line through your eye and your line of sight.
• Angle of depression: you look down at an object. It is the angle between the horizontal through your eye and your line of sight, measured downwards.

The angle of depression of B seen from A equals the angle of elevation of A seen from B. The horizontal line at A and the ground at B are parallel, and the line of sight is a transversal, so they are alternate angles. This lets you move a depression angle down to the ground, where the right angle is.

Choosing the ratio

Two triangles worth memorising

• 30°-60°-90°: sides opposite the angles are in the ratio 1 : √3 : 2. The side opposite 30° is half the hypotenuse.
• 45°-45°-90°: sides are in the ratio 1 : 1 : √2. At 45°, height equals horizontal distance.

With these, many questions need no calculation at all. A pole whose shadow equals its height means the sun is at 45°. A shadow √3 times the height means 30°.

Setups that repeat

One tower, two points on the same side

An observer stands at a point, then walks d metres straight towards a tower of height h. The elevation rises from α to β. Measured from the tower's foot, the far point is h cot α away and the near point is h cot β away. So:

h cot α − h cot β = d

For α = 30° and β = 60°, this is h(√3 − 1/√3) = d, so h = d√3/2.

One tower, points on opposite sides

If the two points are on either side of the tower, the distances add: h cot α + h cot β = d.

The complementary-angle shortcut

If a tower seen from points a and b metres from its foot, on the same line, gives complementary angles of elevation, then h = √(ab). Why: tan α = h/a and tan(90° − α) = cot α = h/b. Multiply them: 1 = h²/(ab).

Heights above the ground

For an object on top of a building (a flagstaff on a tower), or an observer on a building looking at another tower, split the figure into two right triangles sharing the same horizontal distance. Solve for that distance first, then the heights.

Worked questions

Question 1: An observer walks 40 m straight towards a tower. The angle of elevation of the top changes from 30° to 60°. Find the height of the tower and the observer's final distance from it.

• Figure: tower vertical, foot at F. The near point is h cot 60° from F, the far point h cot 30° from F, and the far point is 40 m further out.
• h(√3 − 1/√3) = 40, so h × 2/√3 = 40 and h = 20√3 m (about 34.6 m).
• Final distance = h cot 60° = 20√3 × 1/√3 = 20 m.

Question 2: Two points lie on opposite sides of a 100 m tower, on a straight line through its foot. The angles of elevation of the top from them are 30° and 45°. Find the distance between the points.

• Distances from the foot: 100 cot 30° = 100√3 and 100 cot 45° = 100.
• Total = 100(√3 + 1) m, about 273.2 m.

Question 3: From the top of a 60 m cliff, the angles of depression of two boats in line with the foot, on the same side, are 45° and 30°. How far apart are the boats?

• Move each depression angle to the boat as an elevation angle.
• Near boat: 60 cot 45° = 60 m. Far boat: 60 cot 30° = 60√3 m.
• Gap = 60(√3 − 1) m, about 43.9 m.

Question 4: From the top of a 10 m building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the tower's height.

• Figure: draw a horizontal line from the building's top to the tower. It meets the tower 10 m above the ground.
• Depression 45° to the tower's foot means the horizontal distance equals the 10 m drop, so it is 10 m.
• The part of the tower above that line = 10 tan 60° = 10√3 m.
• Height = 10(1 + √3) m, about 27.3 m.

Question 5: The angles of elevation of a tower's top from two points 4 m and 9 m from its foot, on the same side and in a straight line, are complementary. Find the height.

• h = √(4 × 9) = 6 m.

Question 6: An aeroplane flies horizontally at a height of 1,500√3 m. From a point on the ground, the angle of elevation changes from 60° to 30° in 15 seconds. Find its speed.

• Horizontal distance at 60°: 1,500√3 × 1/√3 = 1,500 m. At 30°: 1,500√3 × √3 = 4,500 m.
• It covers 3,000 m in 15 s, which is 200 m/s = 200 × 18/5 = 720 km/h.

Practice set

1. A 10 m ladder leans against a wall and makes 60° with the ground. How high up the wall does it reach?
2. A kite is flying on a 100 m string that makes 30° with the ground. Find its height.
3. A tower's shadow is √3 times its height. Find the sun's elevation.
4. A tower's shadow becomes 20 m longer when the sun's elevation falls from 60° to 30°. Find the height of the tower.
5. From the top of a 50 m tower, the angle of depression of a car is 30°. How far is the car from the tower's foot?
6. A tower of height 30 m is seen at an elevation of 30° from a point on the ground. How far is the point from the foot?
7. A 6 m flagstaff stands on a tower. From a point on the ground, the angles of elevation of the flagstaff's bottom and top are 30° and 45°. Find the height of the tower.
8. Two poles of equal height stand on either side of an 80 m wide road. From a point on the road between them, the angles of elevation of their tops are 60° and 30°. Find the height of the poles.

Answers

1. 5√3 m. 10 sin 60° = 10 × √3/2.
2. 50 m. 100 sin 30°.
3. 30°. tan θ = 1/√3.
4. 10√3 m. h(√3 − 1/√3) = 20, so h = 20 × √3/2.
5. 50√3 m. 50 cot 30°.
6. 30√3 m. 30 cot 30°.
7. 3(√3 + 1) m, about 8.2 m. Let the distance be x. The tower is x/√3 and tower + 6 = x. So x(1 − 1/√3) = 6, x = 9 + 3√3, and the tower = x/√3 = 3√3 + 3.
8. 20√3 m, about 34.6 m. h cot 60° + h cot 30° = 80, so h × 4/√3 = 80. The point is 20 m from one pole and 60 m from the other.

What to do next

• Redraw every worked question above from its wording alone, without looking at the figure description
• Learn the 1 : √3 : 2 and 1 : 1 : √2 triangles until you can use them without writing
• Revise trigonometry for the value table and identities
• If speed conversions trip you up in aeroplane questions, see speed, time and distance
• Solve the heights and distances questions from the last five CDS papers under time