Heights and Distances for NDA — Angles of Elevation and Depression

Concept

Heights and distances is applied trigonometry — you use ratios of sides in right-angled triangles to find lengths and angles that cannot be measured directly. A surveyor cannot walk up a cliff face to measure it; an air-traffic controller cannot pull a tape measure to a plane. Instead, they measure an angle and a horizontal or slant distance, then let tan, sin, or cos do the rest.

Two angles define everything here:

• Angle of elevation: the angle your line of sight makes with the horizontal when you look upward at an object.
• Angle of depression: the angle your line of sight makes with the horizontal when you look downward at an object.

Here is the analogy that sticks — think of yourself standing on flat ground and holding a horizontal stick at eye level. If you tilt the stick upward to point at a hilltop, the tilt angle is the angle of elevation. If you tilt it downward to point at a boat on the sea below, that tilt is the angle of depression. In both cases, the stick started perfectly horizontal.

One geometric fact you must never forget: the angle of elevation from point A to point B equals the angle of depression from point B to point A. They are alternate interior angles when a transversal cuts two parallel lines (the horizontal at each level). This symmetry rescues you in problems that describe the view from the top of a tower.

The core equation for almost every problem is:

$\tan(\theta) = \frac{\text{perpendicular (vertical height)}}{\text{base (horizontal distance)}}$

When the angle is not 30°, 45°, or 60° — and NDA does throw uncommon angles — you reach for sin or cos directly, and that is when the half-angle and compound-angle formulas become the real tools being tested, not the "heights and distances" concept itself.

Deep Dive

The Basic Setup

Draw the scene before writing a single equation. Every heights-and-distances problem has:

1. One or more observation points (ground level or elevated).
2. One or more objects whose height or distance is unknown.
3. Known angles (of elevation/depression) and at least one known length.

Label all unknowns with single letters. If the tower height is h and the base distance is d, you get one equation per observation point. Two unknowns need two equations — standard system of simultaneous equations.

Standard Angle Values You Must Know Cold

| Angle | sin | cos | tan | |-------|-----|-----|-----| | 30° | 1/2 | √3/2 | 1/√3 | | 45° | 1/√2 | 1/√2 | 1 | | 60° | √3/2 | 1/2 | √3 |

If NDA gives you 22.5°, 67.5°, or 15°, they are testing half-angle formulas — not standard values. Recognise this immediately and switch tools.

Half-Angle Formulas — The NDA Differentiator

These appear in NDA when angles like 67.5° (= 135°/2) show up:

$\sin\frac{\theta}{2} = \sqrt{\frac{1 - \cos\theta}{2}}, \quad \cos\frac{\theta}{2} = \sqrt{\frac{1 + \cos\theta}{2}}$

So sin 67.5° = cos 22.5° (since 67.5° + 22.5° = 90°), and:

$\cos 22.5° = \sqrt{\frac{1 + \cos 45°}{2}} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$

Multi-Point Problems (Collinear Observation)

This is the NDA favourite. Three or more observers on the same horizontal line look at the same tower top. You get one tan equation per observer, all involving the same height h.

Let the tower foot be N, height MN = h. Observers at P, Q, R with angles 30°, 45°, 60°:

$PN = h\cot 30° = h\sqrt{3}, \quad QN = h\cot 45° = h, \quad RN = h\cot 60° = \frac{h}{\sqrt{3}}$

Since P, Q, R, N lie on the same line with P farthest from N:

$PQ = PN - QN = h(\sqrt{3} - 1) = a \implies h = \frac{a}{\sqrt{3}-1} = \frac{a(\sqrt{3}+1)}{2}$

$QR = QN - RN = h\left(1 - \frac{1}{\sqrt{3}}\right) = \frac{h(\sqrt{3}-1)}{\sqrt{3}} = b \implies h = \frac{b\sqrt{3}}{\sqrt{3}-1} = \frac{b(3+\sqrt{3})}{2}$

Then PN = h√3 and MN = h. Substitute and simplify. The rationalisation step (multiply by (√3+1)/(√3+1)) is where most errors happen.

Angle Difference Problems

When you observe two objects from the same point and the question asks for the angle between the two lines of sight (say, from M to the chimney top R and from M to the smoke top Q), use the tangent subtraction formula:

$\tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha \tan\beta}$

This is not just a formula — it is the only efficient route. Never try to work these out geometrically without it.

Bearing and 3D Extensions

NDA occasionally places observers not in a straight line but at known distances from a tower base in different directions. In such cases the "base" of your right triangle is a diagonal of a rectangle or the hypotenuse of a horizontal right triangle. Use Pythagoras first to find the horizontal distance, then apply tan.

Key Structural Moves

• Assign h to the unknown height immediately. Don't name it something verbose.
• Write cot instead of 1/tan when the horizontal distance is the subject — it saves an algebraic inversion step.
• Rationalise early, not at the end. If h = a/(√3 - 1), rationalise it before substituting into PN = h√3. Carrying an irrational denominator through long expressions causes arithmetic errors.
• Check that the geometry is consistent. If R is closer to the tower than Q, RN < QN, so the angle at R must be larger than at Q — if your answer violates this, you've set up the diagram backwards.

Memory Tricks & Shortcuts

Fast-Solving Framework

Read the problem and answer these questions in order:

1. How many observation points? One → single right triangle, one equation. Two or more → system of equations.

2. Are they collinear? Yes → use cot notation, subtract expressions for distances, solve for h. No → find horizontal distances separately using Pythagoras first.

3. Is the angle standard (30°/45°/60°)? Yes → use table values. No → is it half of a standard angle? Yes → half-angle formula. Is it a difference of standard angles? Yes → subtraction formula.

4. What is the question asking for? Height → express h directly. Horizontal distance → use h cot θ. An angle → find both tan values and apply the subtraction formula.

5. Before final answer: Check that larger angles correspond to points closer to the base. Rationalise all surds. Verify the answer has the correct units (km, m — whichever the question uses).

Typical NDA heights-and-distances problem: 3-4 minutes using this flow. Do not start computing before completing step 1-2, or you will set up the wrong number of equations.

Solved PYQs

Why this question: This problem tests whether you recognise 67.5° as a half-angle situation and can apply the half-angle formula for sin (or equivalently cos after complementary angle conversion). Many candidates pick option (d) after an arithmetic slip in the formula.

Solving path: The distance 10 km is the slant distance from the observation point to the plane (not the horizontal ground distance). So height = 10 sin 67.5°. Convert: sin 67.5° = cos 22.5° (complementary angles). Apply the half-angle formula with θ = 45°:

$\cos 22.5° = \sqrt{\frac{1 + \cos 45°}{2}} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{\sqrt{2+\sqrt{2}}}{2}$

Therefore height $= 10 \times \frac{\sqrt{2+\sqrt{2}}}{2} = 5\sqrt{2+\sqrt{2}}$ km. This is option (c).

Why this question: This is a three-point collinear observation problem testing cot manipulation and rationalisation in the same calculation. It appeared as a two-part question in 2025 — Part 1 asks for PN, Part 2 asks for MN. Efficient setup makes both parts trivial.

Solving path: Let MN = h (tower height). Write horizontal distances using cot:

$PN = h\cot 30° = h\sqrt{3}, \quad QN = h\cot 45° = h, \quad RN = h\cot 60° = \frac{h}{\sqrt{3}}$

Since P, Q, R are in order away from N: PQ = PN − QN = h√3 − h = h(√3 − 1) = a.

Rationalise: $h = \frac{a}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{a(\sqrt{3}+1)}{2}$

Then $PN = h\sqrt{3} = \frac{a\sqrt{3}(\sqrt{3}+1)}{2} = \frac{a(3+\sqrt{3})}{2}$. Answer: option (b).

Why this question: The continuation of the same diagram for MN. Having already found h in the previous part, this part requires using the QR = b relation to express the same h differently — the two expressions for h are consistent but the question asks you to express MN in terms of b, not a.

Solving path: From the same diagram: $QR = QN - RN = h - \frac{h}{\sqrt{3}} = h\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right) = b$

So $h = \frac{b\sqrt{3}}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{b\sqrt{3}(\sqrt{3}+1)}{2} = \frac{b(3+\sqrt{3})}{2}$

Therefore $MN = h = \frac{(3+\sqrt{3})b}{2}$. Answer: option (a).

Why this question: This problem layers a chimney problem with a smoke column and asks for the angle between two lines of sight — a pure tangent-subtraction problem disguised as a heights-and-distances scenario. Missing the formula and trying a geometric approach costs 4-5 minutes.

Solving path: Set up coordinates with P at origin, M at (−100, 0) (100 m from base P along the ground). Chimney top R is at (0, 50). Q (smoke top) is directly above P on the line PR extended, and angle QMP = 45°, so PQ = PM = 100, placing Q at (0, 100).

From M: $\tan(\angle QMP) = \frac{PQ}{PM} = \frac{100}{100} = 1$, so $\angle QMP = 45°$. ✓

From M: $\tan(\angle RMP) = \frac{PR}{PM} = \frac{50}{100} = \frac{1}{2}$, so $\angle RMP = \tan^{-1}\!\left(\frac{1}{2}\right)$.

Now $\angle RMQ = \angle QMP - \angle RMP = 45° - \tan^{-1}\!\left(\frac{1}{2}\right)$.

Apply tangent subtraction with $\tan(45°) = 1$ and $\tan\!\left(\tan^{-1}\frac{1}{2}\right) = \frac{1}{2}$:

$\tan(\angle RMQ) = \frac{1 - \frac{1}{2}}{1 + 1 \cdot \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3}$

Therefore $\angle RMQ = \tan^{-1}\!\left(\frac{1}{3}\right)$. Answer: option (b).

Common Mistakes

• Confusing slant distance with horizontal distance. If the problem says "10 km from the observation point" and the angle is the elevation angle, that 10 km is the hypotenuse of the right triangle. Height = 10 sin θ, not 10 tan θ. Always ask: is the given distance horizontal, vertical, or slant?

• Reversing the collinear order. In a tower problem with observers at 30°, 45°, 60°, the observer at 30° is farthest from the tower (smallest angle = largest horizontal distance). Drawing them in the wrong order makes PQ = PN + QN instead of PN − QN, giving a negative or wrong answer.

• Forgetting to rationalise before substituting. Carrying h = a/(√3−1) into subsequent expressions and then trying to rationalise the combined mess is how you make sign errors. Rationalise h as the very next step after finding it.

• Not recognising non-standard angles as half-angles. If you see 22.5°, 67.5°, or 15°, the examiner is testing the half-angle formula. Attempting to find tan 22.5° from scratch without the formula wastes two minutes. Recognise the pattern and move directly to the formula.

• Angle of depression from the wrong horizontal. The angle of depression is measured from the horizontal at the observer's level, not from the vertical or from the ground. A common error is drawing the angle from the vertical, which gives the complement.

• Using the tangent subtraction formula with the wrong order. tan(α − β) assumes α > β. If you compute tan(β − α) with the signs flipped, you get the negative of the correct answer and then cannot match any option. Always identify which angle is larger before applying the formula.

Related Topics

• trigonometric ratios
• trigonometric identities
• inverse trigonometric functions
• half angle formulas
• sine rule cosine rule
• coordinate geometry basics

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