How tall is that tower? How far is the ship from the lighthouse? Questions like these are exactly what Height and Distance is about. In the NDA exam this topic is a gift, because almost every problem reduces to drawing one right-angled triangle and using a single trig ratio. Learn the picture, and you bag 2–4 easy marks every year.
Why Height and Distance Is Easy Marks
Height and Distance is one of the most predictable chapters in NDA Maths. The questions almost never need clever tricks — you draw a right-angled triangle, label the known side and angle, and apply tan, sin or cos. That is it. For a student racing through 120 maths questions, this kind of speed is precious.
The chapter is really just applied trigonometry. Everything you learned about trigonometric ratios in right triangles is now dressed up as a real-world story about towers, hills, kites, ships and aeroplanes. Your only extra job is to translate the words into a clear diagram.
Because the standard angles used are almost always 30°, 45° or 60°, the arithmetic stays clean and the answers come out in tidy forms involving √3. Once you are comfortable with a handful of these values, most questions collapse to two or three lines of working.
Always draw the figure first, even a rough one. A correctly labelled triangle solves three-quarters of the problem before you write a single equation.
The Three Words You Must Know
Every Height and Distance problem is built from just three ideas. Get these crystal clear and the rest is mechanical.
Line of sight
The line of sight is the straight line from the observer's eye to the object being viewed. It is the hypotenuse-or-slant line of your triangle.
Angle of elevation
When the object is above the horizontal level of the eye, the angle between the line of sight and the horizontal is the angle of elevation. You look up, so you raise (elevate) your eyes.
Angle of depression
When the object is below the horizontal level of the eye, the angle between the line of sight and the horizontal is the angle of depression. You look down, so your eyes are depressed.
Both angles are always measured from the horizontal, never from the vertical. Elevation = looking up; depression = looking down.
The Trig Ratios You Will Actually Use
In a right-angled triangle, with respect to an acute angle θ, the three core ratios are defined from the sides. Here "opposite" is the side facing θ, "adjacent" is the side next to θ (not the hypotenuse), and the hypotenuse is opposite the right angle.
sin θ = opposite / hypotenuse
cos θ = adjacent / hypotenuse
tan θ = opposite / adjacent = sin θ / cos θ
In most Height and Distance problems the height is the vertical (opposite) side and the ground distance is the horizontal (adjacent) side. That is why tan θ is by far the most used ratio in this chapter — it links the two sides you usually care about.
The memory hook SOH-CAH-TOA still works: Sin = Opp/Hyp, Cos = Adj/Hyp, Tan = Opp/Adj. Whisper it before every question.
Standard Angle Values to Memorise
Almost every NDA question uses 30°, 45° or 60°. Lock these tiny tables into memory and you will never freeze in the exam.
tan 30° = 1/√3, tan 45° = 1, tan 60° = √3
sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2
cos 30° = √3/2, cos 45° = 1/√2, cos 60° = 1/2
Notice a neat pattern: as the angle grows from 30° to 60°, sin increases while cos decreases, and tan jumps from 1/√3 up to √3. The value √3 ≈ 1.732 turns up constantly, so keep it handy.
tan 45° = 1 means the height equals the horizontal distance. So whenever the angle of elevation is 45°, the object is exactly as far away as it is tall above eye level.
Solving a Single-Triangle Problem
The simplest and most common type gives you one angle and one distance, and asks for the height (or vice versa). Follow a fixed routine and you cannot go wrong.
- Draw the triangle and mark the right angle at the foot of the vertical object.
- Label the known side and the given angle.
- Decide which ratio links what you know to what you want (usually tan).
- Form one equation and solve.
For example, if a tower of height h stands at horizontal distance d from an observer on the ground, and the angle of elevation to the top is θ, then tan θ = h / d. Rearranged, the height is simply h = d × tan θ.
If the observer's eye is at some height (say a person 1.5 m tall), the triangle only gives the height above eye level. Add the eye height back to get the true total height when the question asks for it.
Two-Triangle Problems: When the Observer Moves
Harder NDA questions give two angles — usually because the observer walks closer or further, or there are two observers. These create two right triangles that share the same vertical height.
The standard trick is to call the unknown height h and write a tan equation for each triangle. Both equations contain h, so you can eliminate it and solve for the unknown distance, then back-substitute.
If the angle of elevation of a tower's top changes from α to β as the observer walks a distance d towards the tower (with β > α), the height is:
h = d × (tan α × tan β) / (tan β − tan α)
You do not have to memorise this formula if you can derive it from two tan equations in the exam, but knowing it saves time. The key idea is that the shared height h is the bridge between the two triangles.
When the observer moves towards the object the angle gets bigger; when moving away it gets smaller. Use this to sanity-check your diagram before solving.
Angle of Depression Equals Angle of Elevation
Many students fear depression problems, but there is a beautiful shortcut. The horizontal line at the observer's eye and the horizontal line at the object's level are parallel. The line of sight is a transversal cutting both.
By alternate interior angles, the angle of depression from the top equals the angle of elevation from the bottom. So you can swap one for the other freely.
This means a problem about looking down from a cliff at a boat is mathematically identical to a problem about looking up from the boat at the cliff top. Redraw it as an elevation problem if that feels more comfortable — the numbers stay exactly the same.
Do not mark the angle of depression inside the triangle at the object. It is measured at the observer, between the horizontal and the downward line of sight. The equal angle inside the triangle is the elevation at the object.
Worked Example: The Moving Observer
Let us solve a classic two-triangle problem the way you should in the exam hall.
A man standing on level ground observes the top of a tower at an angle of elevation of 30°. On walking 40 m straight towards the tower, the angle becomes 60°. Find the height of the tower.
The height is 20√3 m ≈ 34.64 m. Notice how writing one tan equation per triangle and eliminating h did all the work.
Common Mistakes to Avoid
Most lost marks here come from careless diagrams and units, not from hard maths.
- Measuring the angle from the vertical instead of the horizontal — always from the horizontal.
- Forgetting to add the observer's eye height when the question wants the full height.
- Mixing up which side is opposite and which is adjacent, so using sin where tan was needed.
- Leaving the answer in mixed units — convert everything to metres (or the asked unit) before computing.
- Using √3 ≈ 1.73 carelessly and rounding too early, which shifts the final answer.
In a moving-observer problem, the 40 m walked is the distance between the two points, not the distance from the tower. Mislabelling this is the single biggest source of wrong answers.
Previous-Year Style Practice
Here is a question modelled on the NDA exam pattern. Try it before reading the solution.
Q. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. What is the height of the tower?
Answer: Here tan 30° = h / 30, so h = 30 × tan 30° = 30 × (1/√3) = 30/√3 = 10√3 m ≈ 17.32 m.
This is the cleanest possible single-triangle problem: one angle, one distance, one tan equation. Rationalising 30/√3 to 10√3 is the only algebra step. Mastering this template makes the harder two-triangle questions feel routine.
Common Problem Shapes in the Exam
NDA examiners recycle a small set of scenarios. Recognise the shape and you already know the method.
Tower and observer
The bread-and-butter case: one vertical tower, one or two ground positions, find the height or the distance using tan.
Two objects on the same line
A flagstaff on top of a building, or a tower with an antenna — here you usually subtract two heights found from two tan equations sharing the same base.
Across a river or valley
The horizontal distance (river width) is unknown; two angles from the two banks let you solve for both width and height.
Aeroplane or balloon overhead
A moving object whose angle changes over a known time or distance; treat it exactly like the moving-observer case.
If a problem mentions a flagstaff on a tower, find the elevation to the top of the flagstaff and to the top of the tower separately, then subtract to get the flagstaff's own height.
Quick Revision Before the Exam
Glance over these the night before your paper and the morning of it.
- Angle of elevation = looking up; angle of depression = looking down; both from the horizontal.
- tan θ = opposite/adjacent = height/distance — your go-to ratio here.
- tan 30° = 1/√3, tan 45° = 1, tan 60° = √3; √3 ≈ 1.732.
- Angle of depression from top = angle of elevation from bottom (parallel lines).
- Two angles → two tan equations sharing height h → eliminate h to solve.
- Always draw and label the triangle first; add eye height if the full height is asked.
Practise 8–10 mixed Height and Distance problems a day for a week. The chapter is small, so a short, focused burst of practice locks in full marks here.
Frequently asked questions
What is the difference between angle of elevation and angle of depression?
The angle of elevation is measured when you look up at an object above your eye level, while the angle of depression is measured when you look down at an object below your eye level. Both are measured from the horizontal line through your eye.
Which trigonometric ratio is used most in Height and Distance?
Tangent (tan) is used most often because it directly relates the vertical height (opposite side) to the horizontal distance (adjacent side). In a typical problem, tan of the angle equals height divided by distance.
Why does the angle of depression equal the angle of elevation?
The horizontal line at the observer and the horizontal line at the object are parallel, and the line of sight is a transversal. By the alternate interior angles rule, the angle of depression from the top equals the angle of elevation from the bottom.
How many questions come from Height and Distance in NDA?
Usually 2 to 4 questions appear in each NDA Maths paper, and they are mostly direct, single-triangle or two-triangle problems. With a little practice this becomes one of the most reliable scoring areas.
Do I need to add the observer's height to the answer?
Only when the question asks for the total height of the object and the observer's eye is at a stated height above the ground. The triangle gives the height above eye level, so you add the eye height to get the true total.
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