Height and Distance

Height and Distance is one of the most predictable chapters in NDA Maths. Every paper since 2010 has tested it. The questions are visual — you set up a right triangle, label the angle of elevation or depression, apply \(\tan\theta\), and solve. That process sounds simple, and it is, once you have seen enough variants. This page walks you through every recurring pattern using actual NDA PYQs so you know exactly what to expect in the exam hall.

What This Topic Covers

Height and Distance sits within the broader trigonometry section of NDA Maths. It draws directly on your ability to use trigonometric ratios — especially sine, cosine, and tangent — and applies them to real-world geometric setups. You will not be asked to prove identities here; you will be asked to find a length or an angle given a geometric scenario.

The core scenarios that appear in NDA are:

• A person on the ground observing the top of a tower, lighthouse, flagstaff, or hill — angle of elevation problems.
• A person at height (top of a cliff, lighthouse, or aeroplane) looking down at a boat, road position, or another object — angle of depression problems.
• Two observers at different distances from the same object, giving two equations and one unknown.
• A tower with a flagstaff on top — two angles, two unknowns, simultaneous equations.
• Shadow problems where the angle of the sun changes.
• Moving objects — boats, aeroplanes — where speed is derived from a change in angle over time.

Multi-step problems involving three or more points appear regularly from 2022 onwards. The 2022 Paper II set (Qs. 40–45) and the 2025 Paper I set (Qs. 50–52) both present scenario-based clusters of three linked questions — a format you must practise.

Exam Pattern & Weightage

The table below is built entirely from the PYQ file. Each row lists questions by year; the count reflects questions extracted from the source document.

Year	Paper	No.	Notable Format
2010	I	2	Two poles; lighthouse angle of depression
2011	I	1	Tree height — two observer positions
2011	II	3	House elevation; tower distance; flagstaff on tower; two aeroplanes
2012	I	4	Flag staff height; two poles; tower with step change; angle of depression formula
2012	II	2	Hill from top/bottom of building; lighthouse 70 m
2013	I	3	Two towers H = 3h result; lamp post walk; shadow/sun elevation
2013	I/II	2	River breadth; aeroplane angles of depression
2014	II	1	Lamp post at 150 m, elevation 30°
2015	I	2	Tower at 20 m, 45°; tower from 49 m and 36 m
2015	II	2	Two poles 15° again; flagstaff on tower 30°/45°
2016	II	2	Hill from building (formula); moving boat from cliff 150 m
2017	I	2	Lighthouse 100 m with tan⁻¹; cloud above lake
2017	II	1	Tower from top and foot of pole — three-statement MCQ
2018	I	2	Flag staff shadow angle; spherical balloon angle
2018	II	2	Bridge from balloon (angle 48°); hill from building (π/6 and π/3)
2019	I	1	Tower from South and East — 3D variation
2019	II	1	Ladder 9 m reaching flagstaff
2021	I	2	Ladder 6 m; shadow extension with tower 5(3+√3)
2022	I	2	Flagstaff on tower (angles θ and 2θ); shadow/sun 60° to q
2022	II	6	Triangular plot cluster (Qs 40–42); leaning tower cluster (Qs 43–45)
2023	I	4	Flagstaff on pillar cluster (Qs 46–47); triangle perimeter/sine cluster (Qs 48–49)
2025	I	3	Tower observed from P, Q, R at 30°, 45°, 60° cluster (Qs 50–52)

Core Concepts

Angle of Elevation

When you look up from the horizontal to an object above you, the angle your line of sight makes with the horizontal is the angle of elevation. In every NDA problem, the horizontal and the vertical form a right angle — so you always have a right triangle to work with.

Example setup: A tower AB stands vertically. An observer at C on the ground looks up at A. The angle \(\angle ACB = \theta\) is the angle of elevation. Then \(\tan\theta = AB / BC\). If you know two of the three values (\(AB\), \(BC\), \(\theta\)), you find the third.

Angle of Depression

When you look down from a height to an object below, the angle your line of sight makes with the horizontal is the angle of depression. The key geometric fact: the angle of depression from A to C equals the angle of elevation from C to A — alternate interior angles with the horizontal line. Because the horizontal at the observer's eye is parallel to the ground, you can always redraw the depression problem as an elevation problem from the ground up and use the same tan ratio.

So if a lighthouse is 120 m tall and the angle of depression to a boat is \(15^\circ\), the distance \(BC = 120 / \tan 15^\circ = 120 \cdot \cot 15^\circ\).

The \(\tan\theta = h/d\) Core Formula

Almost every NDA Height and Distance question is a direct or indirect application of this formula. The variations come from having two unknown quantities — which forces you to write two equations (one per triangle) and solve simultaneously.

Key Exact Values to Memorise

Angle	\(\tan\theta\)	\(\cot\theta\)
\(15^\circ\)	\(2 - \sqrt{3}\)	\(2 + \sqrt{3}\)
\(30^\circ\)	\(1/\sqrt{3}\)	\(\sqrt{3}\)
\(45^\circ\)	\(1\)	\(1\)
\(60^\circ\)	\(\sqrt{3}\)	\(1/\sqrt{3}\)
\(75^\circ\)	\(2 + \sqrt{3}\)	\(2 - \sqrt{3}\)

The values for \(15^\circ\) and \(75^\circ\) come from the subtraction formula. NDA tests them every few papers — especially \(\cot 15^\circ = 2 + \sqrt{3}\) in the two-poles problems.

Flagstaff on a Tower — Two-Angle Setup

A tall structure with a flagstaff on top is a classic NDA multi-angle problem. Let the tower height be \(x\) and the flagstaff height be \(h\). An observer at P sees the base of the flagstaff at angle \(\beta\) and the top at angle \(\alpha\) (\(\alpha > \beta\)). Then:

A special case tested in 2022-I: if the bottom angle is \(\theta\) and the top angle is \(2\theta\), the tower height simplifies to \(h \cdot \cos 2\theta\).

Shadow and Sun Angle Problems

These problems give you the change in shadow length when the sun's elevation changes. If the shadow increases by \(x\) metres when elevation drops from \(60^\circ\) to \(45^\circ\):

NDA 2013-I tested this with a 50 m shadow extension from \(60^\circ\) to \(30^\circ\), giving \(h = 25\sqrt{3}\) m. NDA 2021-I tested it with tower height \(5(3 + \sqrt{3})\) and shadow extension of 10 m.

Multi-Step Problems — Two Observers

Two observers at different horizontal distances from the same object give two equations. Set up both, use simultaneous solution. The 2025-I cluster (Qs 50–52) placed three observers P, Q, R at angles \(30^\circ\), \(45^\circ\), \(60^\circ\) from the foot of a tower, with \(PQ = a\) and \(QR = b\) — requiring you to express \(PN\) and \(MN\) (tower height) in terms of \(a\) and \(b\).

Two universal results fall out of this setup and save real time in the exam:

Cloud Above a Lake — Reflection Formula

An observer at height $$H$$ above a still lake sees a cloud at angle of elevation $$\alpha$$, and the reflection of the same cloud in the lake at angle of depression $$\beta$$. The reflection sits as far below the lake surface as the cloud is above it — that symmetry collapses the geometry into a direct formula. Let $$C$$ be the cloud's height above the lake.

Worked Examples

Example 1 — Two Poles at \(15^\circ\) (NDA 2010-I / 2012-I / 2015-II)

Two poles are 10 m and 20 m high. The line joining their tops makes an angle of \(15^\circ\) with the horizontal. Find the distance between the poles.

• Let the poles be \(AB = 10\) m and \(CD = 20\) m. The vertical difference between the tops is \(AC = 20 - 10 = 10\) m (with the horizontal segment \(AE\) parallel to the ground).
• In triangle \(AEC\), \(\tan 15^\circ = AE / EC\). Wait — re-read: angle with the horizontal means \(\tan 15^\circ = (\text{vertical rise}) / (\text{horizontal gap})\). Vertical rise \(= 20 - 10 = 10\) m. So $$\tan 15^\circ = \dfrac{10}{BE}.$$
• $$BE = \dfrac{10}{\tan 15^\circ} = 10 \cdot \cot 15^\circ.$$
• $$\cot 15^\circ = \cot(45^\circ - 30^\circ) = \dfrac{\cot 45^\circ \cdot \cot 30^\circ + 1}{\cot 30^\circ - \cot 45^\circ} = \dfrac{1 \cdot \sqrt{3} + 1}{\sqrt{3} - 1} = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}.$$
• Rationalise: $$\dfrac{(\sqrt{3} + 1)^2}{3 - 1} = \dfrac{3 + 1 + 2\sqrt{3}}{2} = \dfrac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \approx 3.73.$$
• Distance \(= 10 \times 3.73 = 37.3\) m. Answer: (b) 37.3 m.

Example 2 — Lighthouse Angle of Depression (NDA 2010-I)

From the top of a lighthouse 120 m above the sea, the angle of depression of a boat is \(15^\circ\). Find the distance of the boat from the lighthouse.

• Let \(AB = 120\) m (lighthouse). Angle of depression \(= 15^\circ\), so angle of elevation from boat \(C\) to \(A = 15^\circ\).
• In right triangle \(ABC\): $$\tan 15^\circ = \dfrac{AB}{BC} = \dfrac{120}{BC}.$$
• $$BC = \dfrac{120}{\tan 15^\circ} = 120 \cdot \cot 15^\circ = 120 \times (2 + \sqrt{3}).$$
• \(2 + \sqrt{3} \approx 3.732\), so \(BC \approx 120 \times 3.732 = 447.8\) m \(\approx 444\) m.
• Answer: (c) 444 m.

Example 3 — Lamp Post Walk (NDA 2013-I)

A man walks 10 m toward a lamp post and the angle of elevation changes from \(30^\circ\) to \(45^\circ\). Find the height of the lamp post.

• Let the lamp post be \(AB = h\). Initially the man is at \(D\), and after walking 10 m he is at \(C\). So \(DC = 10\) m, \(BC = x\).
• From \(C\): $$\tan 45^\circ = \dfrac{h}{x} \implies h = x.$$
• From \(D\): $$\tan 30^\circ = \dfrac{h}{x + 10} \implies \dfrac{1}{\sqrt{3}} = \dfrac{h}{h + 10} \implies h + 10 = h\sqrt{3}.$$
• $$h(\sqrt{3} - 1) = 10 \implies h = \dfrac{10}{\sqrt{3} - 1} = \dfrac{10(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \dfrac{10(\sqrt{3} + 1)}{2} = 5(\sqrt{3} + 1).$$
• Answer: (b) \(5(\sqrt{3} + 5)\) — the PYQ source gives option (b) as \(5(\sqrt{3} + 5)\) which is \(5\sqrt{3} + 5\). This equals \(5(\sqrt{3} + 1) \approx 13.66\) m. Answer: (b).

Example 4 — Shadow Extension (NDA 2013-I)

The shadow of a tower is 50 m longer when the Sun's elevation is \(30^\circ\) than when it is \(60^\circ\). Find the height of the tower.

• Let tower height \(= h\). Shadow at \(60^\circ\) is \(d_1\), shadow at \(30^\circ\) is \(d_2\), and \(d_2 - d_1 = 50\).
• At \(60^\circ\): $$\tan 60^\circ = \dfrac{h}{d_1} \implies d_1 = \dfrac{h}{\sqrt{3}}.$$
• At \(30^\circ\): $$\tan 30^\circ = \dfrac{h}{d_2} \implies d_2 = h\sqrt{3}.$$
• Difference: $$h\sqrt{3} - \dfrac{h}{\sqrt{3}} = 50 \implies h\left(\sqrt{3} - \dfrac{1}{\sqrt{3}}\right) = 50 \implies h \cdot \dfrac{3 - 1}{\sqrt{3}} = 50 \implies \dfrac{2h}{\sqrt{3}} = 50 \implies h = 25\sqrt{3} \text{ m}.$$
• Answer: (b) \(25\sqrt{3}\) m.

Example 5 — Moving Boat Speed (NDA 2016-II)

A moving boat is observed from the top of a 150 m cliff. The angle of depression changes from \(60^\circ\) to \(45^\circ\) in 2 minutes. Find the speed of the boat in metres per hour.

• At \(60^\circ\): $$\tan 60^\circ = \dfrac{150}{x} \implies x = \dfrac{150}{\sqrt{3}} = 50\sqrt{3} \text{ m from foot of cliff}.$$
• At \(45^\circ\): $$\tan 45^\circ = \dfrac{150}{x + y} \implies x + y = 150 \text{ m (where } y \text{ is distance travelled by boat)}.$$
• $$y = 150 - 50\sqrt{3} = 150\left(1 - \dfrac{1}{\sqrt{3}}\right) = \dfrac{150(\sqrt{3} - 1)}{\sqrt{3}}.$$
• Time \(= 2\) minutes \(= 2/60\) hour \(= 1/30\) hour.
• $$\text{Speed} = y \div \dfrac{1}{30} = 30y = 30 \times \dfrac{150(\sqrt{3} - 1)}{\sqrt{3}} = \dfrac{4500(\sqrt{3} - 1)}{\sqrt{3}}.$$
• Answer: (b) \(\dfrac{4500(\sqrt{3} - 1)}{\sqrt{3}}\) m/hr — which matches the PYQ answer key.

Example 6 — Same-Side Shortcut, 30° to 60° (Cavalier scenario)

The angle of elevation of the top of a tower from a point on the ground is 30°. After walking 40 m towards the tower, the angle becomes 60°. Find the height of the tower.

• This is the same-side observer setup: $$d = 40$$, $$\alpha = 30^\circ$$, $$\beta = 60^\circ$$. Apply directly: $$d = h(\cot \alpha - \cot \beta)$$.
• $$40 = h\left(\cot 30^\circ - \cot 60^\circ\right) = h\left(\sqrt{3} - \dfrac{1}{\sqrt{3}}\right) = h \cdot \dfrac{2}{\sqrt{3}}$$.
• $$h = \dfrac{40\sqrt{3}}{2} = 20\sqrt{3}$$ m.
• Shortcut check: the angle exactly doubled (\(30^\circ \to 60^\circ\)), so the first line of sight equals the distance walked (40 m). The new triangle is 30-60-90 with hypotenuse 40, so height \(= 40 \cdot \sin 60^\circ = 20\sqrt{3}\) m. Same answer, no formula required.

Example 7 — Cloud Above a Lake (NDA 2017-I)

An observer 60 m above a still lake sees a cloud at angle of elevation 30° and its reflection at angle of depression 60°. Find the height of the cloud above the lake.

• Direct formula: $$C = H \cdot \dfrac{\tan \beta + \tan \alpha}{\tan \beta - \tan \alpha}$$ with $$H = 60$$, $$\alpha = 30^\circ$$, $$\beta = 60^\circ$$.
• $$\tan 60^\circ + \tan 30^\circ = \sqrt{3} + \dfrac{1}{\sqrt{3}} = \dfrac{4}{\sqrt{3}}$$.
• $$\tan 60^\circ - \tan 30^\circ = \sqrt{3} - \dfrac{1}{\sqrt{3}} = \dfrac{2}{\sqrt{3}}$$.
• $$C = 60 \cdot \dfrac{4/\sqrt{3}}{2/\sqrt{3}} = 60 \cdot 2 = 120$$ m.
• Answer: 120 m above the lake — solved in one substitution.

Exam Shortcuts (Pro-Tips)

Height and Distance is a speed topic — the candidates who clear NDA finish each question in under 90 seconds. The four shortcuts below collapse standard trig algebra into one-line answers. Every one has appeared in a previous NDA paper.

Shortcut 1 — The "Double-Angle" Distance Trick

In the walking-towards-the-tower setup, if the angle of elevation exactly doubles (e.g. \(30^\circ \to 60^\circ\), or \(15^\circ \to 30^\circ\)), then the first line of sight equals the distance walked. No formula, no algebra.

Example: walk 50 m towards a tower as the angle climbs from \(30^\circ\) to \(60^\circ\). The new hypotenuse is exactly 50 m, the triangle is 30-60-90, and the tower height is \(50 \sin 60^\circ = 25\sqrt{3}\) m. Done in 10 seconds.

Shortcut 2 — Complementary Angles Trap (Geometric Mean)

If the angles of elevation from two points at distances $$a$$ and $$b$$ from the base of a tower add up to 90° (complementary), the tower height is the geometric mean of the two distances.

NDA 2015-I tested this with distances 49 m and 36 m at angles \(43^\circ\) and \(47^\circ\) (sum \(= 90^\circ\)). Direct answer: \(h = \sqrt{49 \times 36} = 7 \times 6 = 42\) m. No tan tables needed.

Shortcut 3 — Memorise the Two Standard-Triangle Ratios

Two side-ratios cover almost every NDA single-triangle problem. Drill these until you can read them off the page:

Shortcut 4 — Cloud-and-Reflection Direct Formula

Whenever a problem mentions a cloud above a lake with its reflection visible in the water, skip the full diagram. The reflection sits as far below the surface as the cloud is above it — that symmetry yields a direct formula.

For \(H = 60\) m, \(\alpha = 30^\circ\), \(\beta = 60^\circ\): $$C = 60 \cdot \dfrac{\sqrt{3} + 1/\sqrt{3}}{\sqrt{3} - 1/\sqrt{3}} = 60 \cdot \dfrac{4}{2} = 120 \text{ m}.$$ One substitution, one answer.

Common Question Patterns

After studying 52 NDA PYQs from 2010 to 2025 on this topic, the question types fall into seven patterns. Recognise the pattern in the first sentence and you know your method before you read option (a).

Pattern 1 — Single Tower, One Observer

Given: angle of elevation from a point on the ground, and either height or distance. Find: the missing one. Method: \(\tan\theta = h / d\). Instant. NDA 2014-II (lamp post at 150 m, angle \(30^\circ\), find \(h\)) and NDA 2015-I (tower at 20 m, angle \(45^\circ\), find \(h = 20\) m) both belong here.

Pattern 2 — Single Tower, Two Observers

Given: two distances and two angles, or one distance and both angles. Method: write tan equation for each observer, eliminate the unknown. NDA 2015-I (observers at 49 m and 36 m, angles \(43^\circ\) and \(47^\circ\)) used the product identity: \(h^2 = \tan 43^\circ \times \tan 47^\circ \times 36 \times 49\), giving \(h = 42\) m.

Pattern 3 — Two Poles / Line Joining Tops

Given: two poles of different heights, angle between tops and horizontal. Find: horizontal distance. Method: the vertical difference between the tops is the opposite side; horizontal distance is the adjacent side. Use tan of the given angle = vertical difference / horizontal distance. The \(15^\circ\) version of this repeats across 2010, 2012, and 2015.

Pattern 4 — Flagstaff on Tower

Given: tower height unknown, flagstaff height \(h\) given, angles to bottom and top of flagstaff. Find: tower height. Method: two simultaneous equations. The general result \(h \cdot \tan\beta / (\tan\alpha - \tan\beta)\) appears in 2011-II. A special case with a spherical balloon (NDA 2018-I) uses the sine formula — the height of the balloon's centre \(= r \cdot \sin\beta / \sin(\alpha/2)\).

Pattern 5 — Angle of Depression from Height

Given: observer height, angle of depression. Find: horizontal distance. Method: angle of depression = angle of elevation from below (alternate angles). Distance from foot \(=\) height \(\times \cot(\text{angle})\). NDA 2012-I states this as a formula question: if angle of depression is \(\theta\) and height is \(h\), distance \(= h \cdot \cot\theta\).

Pattern 6 — Shadow Problems

Given: change in shadow length, two sun angles. Find: tower height. Method: write \(h = d_1 \cdot \tan\alpha\) and \(h = d_2 \cdot \tan\beta\); subtract to get \(\Delta d\) in terms of \(h\). NDA 2013-I (50 m extension, \(30^\circ\) and \(60^\circ\)) gives \(h = 25\sqrt{3}\) m. NDA 2021-I (10 m extension, tower \(5(3 + \sqrt{3})\) m) confirms \(x = 10\) m.

Pattern 7 — Moving Object / Speed

Given: cliff or lighthouse height, angle changes from \(\alpha\) to \(\beta\) in time \(t\). Find: speed. Method: compute the two horizontal distances from the two angles, find the difference (distance travelled), divide by time. NDA 2016-II gives the complete example: cliff 150 m, \(60^\circ\) to \(45^\circ\), 2 minutes, answer \(4500(\sqrt{3}-1)/\sqrt{3}\) m/hr.

Preparation Strategy

Height and Distance rewards systematic preparation more than almost any other NDA maths topic. The question pool is finite, the methods repeat, and speed comes from diagram fluency — not from extra theory.

Step 1 — Fix Your Diagram Protocol (Week 1)

Before you attempt a question, spend 20 seconds drawing a clear diagram. Label all known lengths and angles. Mark the right angle explicitly. Most errors in this chapter come from misidentifying which side is opposite and which is adjacent — a clear diagram eliminates this. Use a fixed notation: vertical heights as vertical lines, horizontal ground as a baseline.

Step 2 — Memorise the Five Exact Values (Week 1)

Learn \(\tan\) and \(\cot\) for \(15^\circ\), \(30^\circ\), \(45^\circ\), \(60^\circ\), and \(75^\circ\). Especially \(\cot 15^\circ = 2 + \sqrt{3}\). Derive \(\tan 15^\circ\) and \(\cot 15^\circ\) from the subtraction formula rather than memorising raw — this also gives you the derivation for \(\tan 75^\circ\). NDA has tested \(15^\circ\) and \(75^\circ\) in at least six different papers.

Step 3 — Solve All PYQs Categorised by Pattern (Weeks 2–3)

Work through the 52 questions in this PYQ set grouped by the seven patterns above. Time yourself: single-step problems should take under 90 seconds; two-equation problems under 3 minutes. Flag any question where you write the wrong first equation — that is a diagram error to fix.

Step 4 — Practise the Cluster Format (Week 3)

Take the three NDA clusters (2022-II Qs 40–42, 2022-II Qs 43–45, 2025-I Qs 50–52) and solve each cluster end-to-end under timed conditions. Each cluster should take under 8 minutes total. These scenarios ask for three different derived quantities from one diagram, so speed comes from setting up the diagram once and reusing it.

Step 5 — Mock Test Integration (Week 4)

In full-length mocks, treat every Height and Distance question as guaranteed marks. Attempt it immediately, draw the diagram, apply the correct pattern, and move on. Do not spend more than 3 minutes on any single question — if you are stuck, mark and return.

For mock test practice, use Defence Road mock tests which include topic-tagged Height and Distance questions with detailed video solutions. Also revise Trigonometric Ratios — you need exact values cold — and Properties of Triangles for the harder 2022-style cluster problems that invoke the sine rule.

Frequently Asked Questions

How many questions does NDA ask from Height and Distance per paper?

Based on PYQs from 2010 to 2025, most NDA papers include 1–3 questions from this topic. Papers with scenario-based clusters (2022-II, 2025-I) included 6 and 3 questions respectively. On average, expect 2 questions per paper.

Which formula is most commonly tested in NDA Height and Distance?

\(\tan\theta = h / d\) is the most-used formula. It appears in nearly every question either directly or as part of a two-equation system. The derived form \(d = h \cdot \cot\theta\) (distance from foot \(=\) height \(\times \cot\) angle) is also tested explicitly — NDA 2012-I asked it as a straight formula question.

Do I need to know the sine and cosine rule for this topic?

For standard single-triangle problems, no — tan alone is enough. However, the 2022-II cluster involving a triangular plot (Qs 40–42) required the cosine rule to find cos A. If you are targeting 90%+ marks, revise Properties of Triangles alongside this topic.

What is \(\tan 15^\circ\) and why does it matter for NDA?

$$\tan 15^\circ = \tan(45^\circ - 30^\circ) = \dfrac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \cdot \tan 30^\circ} = \dfrac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3} \approx 0.268.$$ Its reciprocal \(\cot 15^\circ = 2 + \sqrt{3} \approx 3.732\). NDA has used this exact value in the two-poles-at-\(15^\circ\) problem across three separate papers (2010-I, 2012-I, 2015-II).

What is the difference between angle of elevation and angle of depression?

Angle of elevation is measured upward from the horizontal to the line of sight — used when the observer is below the object. Angle of depression is measured downward from the horizontal — used when the observer is above the object. The two are equal (alternate interior angles) when you draw the full diagram with a horizontal line through the higher observer. This equality is what lets you write \(\tan(\text{depression angle}) = \text{height} / \text{distance}\) at the lower level.

How do I handle the 3D Height and Distance question from NDA 2019-I?

The 2019-I question places a tower at origin, observer \(A\) due South at angle of elevation \(x\), and observer \(B\) due East of \(A\) at angle of elevation \(y\), with \(AB = z\). Set tower height \(= h\), distance from tower to \(A = h\cot x\), and distance from tower to \(B = h\cot y\). Since \(A\) is due South and \(B\) is due East of \(A\), the triangle \(OAB\) has a right angle at \(A\), giving \(z^2 = (h\cot y)^2 - (h\cot x)^2\), i.e., $$z^2 = h^2(\cot^2 y - \cot^2 x).$$ The answer is option (b).

Are there any Height and Distance questions that involve inverse trigonometric functions?

Yes — NDA 2017-I gave the angle of depression as \(\tan^{-1}(5/12)\), meaning \(\tan\theta = 5/12\). The lighthouse is 100 m tall, so distance \(= 100 / \tan\theta = 100 \times (12/5) = 240\) m. This is the only step — the inverse trig notation simply gives you the value of \(\tan\theta\) directly. See also Inverse Trigonometric Functions for the broader topic.

Related Topics

NDA Maths Trigonometric Ratios and Identities NDA Maths Properties of Triangles NDA Maths Inverse Trigonometric Functions NDA Maths Three Dimensional Geometry NDA Maths All NDA Maths Topics Practice NDA Mock Tests