Height and Distance

Why Height and Distance Is Easy Marks

In the CDS Elementary Mathematics paper you can expect 2 to 4 questions from Height and Distance almost every year. The topic looks like real-world measurement — the height of a tower, the distance of a ship, the breadth of a river, the length of a shadow — but underneath it is pure right-angled trigonometry. There is no theorem to prove and no long derivation to remember; you simply translate the words into a triangle and read off the answer.

The reason it is so reliable is that the examiner uses only three angles: 30°, 45° and 60°. Once you memorise the ratios for these angles and learn to draw a clean diagram, the algebra is short. No calculus, no logarithms, no long formulas — just one or two triangles and the tangent ratio. Because the working is so mechanical, this is one of the few topics where a well-prepared candidate can guarantee full marks rather than hoping for them.

Treat Height and Distance as a speed booster in the paper. The minute you save here by answering instantly is a minute you can spend on a harder mensuration or algebra question. Many toppers attempt these first to build confidence and bank easy marks early.

Angle of Elevation and Angle of Depression

Both angles are always measured from the horizontal line through the observer's eye, never from the vertical.

Angle of elevation

When you look up at an object higher than your eye, the angle between your line of sight and the horizontal is the angle of elevation. A short person looking at a tall tower top sees a large elevation angle.

Angle of depression

When you look down at an object below your eye — a boat from a cliff or a car from a tower, for example — the angle between the horizontal and your downward line of sight is the angle of depression. A crucial point: the higher you stand, the larger the angle of depression to a fixed object below.

Students often confuse the two because the words sound technical, but the rule is simple. Elevation means your eyes go up; depression means your eyes go down. Both are measured from an imaginary horizontal line drawn through your eye, and that horizontal is the reference for every single problem in this chapter.

The Right Triangle and Its Three Ratios

Every Height and Distance figure contains a right-angled triangle. Name the sides relative to the angle θ you are using:

• Opposite — the side facing θ (usually the height).
• Adjacent — the side beside θ along the ground (usually the distance).
• Hypotenuse — the slant line of sight, opposite the right angle.

A handy memory aid is the phrase SOH-CAH-TOA: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent. Drill this until it is automatic.

In nearly all CDS problems you know the ground distance and need the height, or you know the height and need the distance — so tangent is the workhorse ratio, because it links exactly those two sides. Reach for sin or cos only when the slant distance (the hypotenuse, often a rope, a ladder or a line of sight) is given or asked for.

Standard Angle Values You Must Memorise

These are the only values the examiner will use. Learn them cold.

Notice the symmetry: tan 30° and tan 60° are reciprocals (1÷√3 and √3). This is the basis of a very common shortcut you will see below.

Solving a Single-Triangle Problem

The simplest type gives one angle and one distance. The method is fixed:

1. Draw the right triangle and label the height h and the ground distance d.
2. Write tan(angle) = h ÷ d.
3. Substitute the standard value and solve for the unknown.

If instead you are told the height and the angle and asked for the distance, just rearrange the same equation: d = h ÷ tan(angle). The triangle never changes; only the unknown moves. This is why the topic feels so light once you have practised a dozen sums — the structure is identical every time.

A classic single-triangle case is the shadow problem. A vertical pole casts a shadow on level ground, and the angle of elevation of the Sun is the angle at the tip of the shadow. If the Sun's elevation is 45°, the shadow length equals the pole height. As the Sun climbs higher the shadow shortens; as it sinks towards sunset the shadow lengthens. Knowing this behaviour lets you sanity-check your answer in a glance.

Worked Example: Height of a Tower

So the tower is about 34.6 m tall. Note how rationalising 60÷√3 gives the neat exact form 20√3 m — CDS options are often written in surd form, so keep both the decimal and the surd handy.

Two-Triangle Problems and the Common Side

Harder questions involve two angles observed from two points, or one observer who walks closer to the object. These give two right triangles that share a common side — usually the height.

The standard set-up

An observer sees the top of a tower at angle 30°. Walking a metres nearer, the angle becomes 60°. Let the height be h and the nearer distance be x. Then:

• From the near point: tan 60° = h ÷ x, so h = x√3.
• From the far point: tan 30° = h ÷ (x + a), so h = (x + a) ÷ √3.

Equate the two expressions for h and solve for x, then back-substitute to get the height. The whole method rests on the fact that both triangles climb to the same top point, so their vertical side is identical.

The same idea covers problems where the two observations come from different vantage points — a man on the bank and a man on a bridge, or a balloon seen from two stations. As long as you can identify the shared side and write a tangent equation for each angle, you reduce two pieces of information to two equations in two unknowns and solve them like any ordinary pair of simultaneous equations.

Ready Shortcuts for the Classic Cases

For the walk-closer problem with angles 30° and 60° and walked distance a, the results fall out cleanly:

So if the observer walks 40 m and the angle rises from 30° to 60°, the tower is 40√3÷2 = 20√3 ≈ 34.6 m and the foot is 20 m from the near point. Deriving the formula once is fine, but memorising it saves a minute under exam pressure.

Why does the geometric-mean shortcut work? If the elevation is θ from one point and (90° − θ) from the other, then tanθ = h÷d₁ and tan(90° − θ) = cotθ = h÷d₂. Multiplying the two equations gives tanθ × cotθ = h² ÷ (d₁ × d₂). Since tanθ × cotθ = 1, we get h² = d₁ × d₂, so h = √(d₁ × d₂). Spotting the word “complementary” or a pair like 30° and 60° is your cue to apply it directly and skip the algebra.

Build a small library of these standard outcomes during revision — the √3 shadow giving 30°, the equal shadow giving 45°, the walk-closer formula, and the geometric-mean rule. In the exam you will recognise the pattern within seconds and simply substitute the numbers, which is exactly the kind of speed the CDS Maths paper rewards.

Common Mistakes That Cost Marks

One more: when a problem mentions a ship and a lighthouse, the angle of depression is given at the top, but the right triangle's vertical side is still the full lighthouse height. Redraw using alternate angles to avoid confusion.

Previous-Year Style Question

A Quick Method Checklist

Use this fixed sequence on every Height and Distance question to stay fast and accurate:

1. Read once and draw the right triangle; mark the right angle.
2. Label every known angle and side; identify the unknown.
3. Choose the ratio: tan for height-and-ground problems, sin/cos when the slant line is involved.
4. Substitute the standard value (30°, 45° or 60°) and solve.
5. Add any observer height; check the answer against the options in surd and decimal form.

Quick Revision