Heights and distances for NDA

Heights and distances is trigonometry with a picture attached, and the picture is where the marks are won or lost. The calculation is almost always one or two tangents of 30°, 45° or 60°. What goes wrong is the diagram: the angle placed at the wrong corner, the observer's height ignored, or two triangles merged into one. Draw first, label every length, and the algebra becomes routine.

Typical question types are:

• the height of a tower or the distance of a point from one angle and one length;
• two observation points on the same side of a tower, or on opposite sides;
• a flagstaff or statue standing on a building;
• angles of depression from a cliff, lighthouse or aeroplane, often with a boat moving;
• shadows that lengthen as the sun's elevation falls;
• ladders, kite strings and broken trees, where the slant length is given.

The vocabulary

The line of sight runs from the observer's eye to the object. The angle of elevation is the angle the line of sight makes with the horizontal when you look up. The angle of depression is the angle it makes with the horizontal when you look down.

Because the horizontal through the observer and the ground are parallel, the angle of depression from the top equals the angle of elevation from the bottom (alternate angles). This one fact converts every depression problem into an ordinary right triangle drawn at ground level.

Unless the question says otherwise, towers are vertical, the ground is level, and the observer is a point on the ground.

The method

1. Draw the vertical object, the horizontal ground and the line of sight.
2. Mark the angle at the observer, not at the top of the tower.
3. Label the unknown height h and the unknown distance x.
4. Use tan = opposite/adjacent when height and horizontal distance are involved. Use sin or cos only when a slant length (ladder, string, line of sight) is given.
5. With two angles, write one equation per right triangle and eliminate x.

Values to have ready: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3, tan 15° = 2 − √3, tan 75° = 2 + √3, √3 ≈ 1.732.

Shortcut results

Derive these once so you trust them, then use them directly.

Where the first one comes from. If the nearer point is x from the foot, h = x tan β and h = (x + d) tan α. So x = h cot β and h cot α − h cot β = d, which rearranges to the formula. The third row is quicker still: tan θ = h/a and cot θ = h/b, and multiplying gives 1 = h²/(ab).

Worked NDA-style MCQs

Q1. From two points on the same side of a tower and 20 m apart, the angles of elevation of its top are 30° and 60°. The height of the tower is:
(a) 10 m (b) 10√3 m (c) 20√3 m (d) 20 m

Let the nearer point be x m away. Then h = x√3 and h = (x + 20)/√3. So 3x = x + 20, x = 10 and h = 10√3 m. The shortcut agrees: 20 × (1/√3)(√3)/(√3 − 1/√3) = 20/(2/√3) = 10√3. Answer: (b).

Q2. The angles of elevation of the top of a tower from two points 4 m and 9 m from its base, on the same straight line, are complementary. The height of the tower is:
(a) 5 m (b) 6 m (c) 6.5 m (d) 13 m

h = √(4 × 9) = 6 m. Answer: (b).

Q3. From a point on the ground, the angles of elevation of the bottom and top of a flagstaff standing on a 20 m building are 45° and 60°. The length of the flagstaff is:
(a) 20 m (b) 20√3 m (c) 20(√3 − 1) m (d) 10(√3 − 1) m

The 45° angle to the bottom means the point is 20 m from the building. The top is then at 20 tan 60° = 20√3 m. The flagstaff is 20√3 − 20 = 20(√3 − 1) ≈ 14.6 m. Answer: (c).

Q4. From the top of a 60 m cliff, the angles of depression of the top and bottom of a tower are 30° and 60°. The height of the tower is:
(a) 20 m (b) 30 m (c) 40 m (d) 45 m

The bottom is at depression 60°, so the horizontal distance is 60/tan 60° = 60/√3 = 20√3 m. The top of the tower is at depression 30°, so it is 20√3 × tan 30° = 20 m below the cliff top. The tower is 60 − 20 = 40 m. Answer: (c).

Q5. From the top of a 75 m lighthouse, the angle of depression of a ship changes from 30° to 45° as the ship sails straight towards it. The distance travelled is:
(a) 75 m (b) 75(√3 − 1) m (c) 75(√3 + 1) m (d) 75√3 m

Initial distance 75 cot 30° = 75√3. Final distance 75 cot 45° = 75. The ship travels 75(√3 − 1) ≈ 54.9 m. Answer: (b).

Q6. A person 1.5 m tall stands 28.5 m from a chimney and sees its top at an elevation of 45°. The height of the chimney is:
(a) 28.5 m (b) 30 m (c) 27 m (d) 29 m

The 45° triangle starts at eye level: the part above the eye is 28.5 m. Add the eye height: 28.5 + 1.5 = 30 m. Answer: (b).

Common mistakes

• Putting the angle of depression inside the triangle at the top. Move it to the observer on the ground as an angle of elevation.
• Ignoring the observer's height when it is given. It is there to be added.
• Using sin when the horizontal distance is given. Height and horizontal distance always pair with tan.
• Mixing up which point is nearer. The nearer point has the larger angle of elevation.
• Rounding √3 too early. Keep surds until the end; the options usually do.

Practice set

1. A kite string 100 m long makes 30° with the ground. The height of the kite is: (a) 50√3 m (b) 50 m (c) 100 m (d) 100/√3 m
2. A 6 m pole stands in sunlight at an elevation of 30°. Its shadow is: (a) 6√3 m (b) 2√3 m (c) 3 m (d) 12 m
3. From the top of a 40 m tower, the angle of depression of a point on the ground is 45°. The point is: (a) 20 m away (b) 40 m away (c) 40√3 m away (d) 80 m away
4. The shadow of a tower is √3 times its height. The sun's elevation is: (a) 60° (b) 45° (c) 30° (d) 15°
5. A tower's shadow is 40 m longer when the sun's elevation is 30° than when it is 60°. The height of the tower is: (a) 20 m (b) 20√3 m (c) 40√3 m (d) 40 m
6. A ladder whose foot is 2.5 m from a wall makes 60° with the ground. Its length is: (a) 5 m (b) 2.5√3 m (c) 5√3 m (d) 10 m
7. A tree breaks, and its top touches the ground 10 m from the foot, making 30° with the ground. The original height was: (a) 10 m (b) 20 m (c) 10√3 m (d) 20√3 m
8. The elevations of a tower's top from points 9 m and 16 m from its foot, on the same line, are complementary. The height is: (a) 12 m (b) 12.5 m (c) 25 m (d) 7 m

Answers:

1. (b). 100 sin 30° = 50 m.
2. (a). 6/tan 30° = 6√3 m.
3. (b). tan 45° = 1, so the distance equals the height.
4. (c). tan θ = h/(h√3) = 1/√3.
5. (b). h(cot 30° − cot 60°) = 40, so h(√3 − 1/√3) = 40, h × 2/√3 = 40 and h = 20√3 m.
6. (a). 2.5/cos 60° = 5 m.
7. (c). Standing part 10 tan 30° = 10/√3; broken part 10/cos 30° = 20/√3. Total 30/√3 = 10√3 m.
8. (a). √(9 × 16) = 12 m.

What to do next

• Re-derive the four shortcut results from diagrams, without looking.
• Solve 25 old NDA heights-and-distances questions, drawing every figure, even the easy ones.
• Revise the ratios and allied angles in trigonometric identities. For triangles that are not right-angled, use the sine and cosine rules in properties of triangles.