NDA Maths · Height & Distance

Shadows, Leaning Structures & Special Geometry

When the sun's elevation changes, a tower's shadow stretches or shrinks; when a tower leans, its top no longer sits over its foot; and a few questions hide a chord or arc of a circle. All three are still right-triangle reasoning, just with one extra twist.

All Height & Distance notes NDA Maths strategy

Why this matters

A smaller but punishing subtopic: 8 PYQs, 6 of them HARD. The shadow problems test whether you can read the sun's elevation as the angle in the height-over-shadow triangle. The leaning-tower set is the chapter's hardest cluster — a structure tilted off the vertical needs two elevation readings to separate its true height from its lean. A couple of questions are really circle geometry (chord length, arc length) wearing a height-and-distance label. Recognise which is which and each becomes routine.

Concept 1 of 5

Shadows and the Sun's Elevation

Intuition

The sun's rays hit the ground at the angle of elevation of the sun, and a vertical tower casts a shadow exactly as long as the base of that right triangle. A lower sun (smaller elevation) casts a longer shadow. The height stays fixed, so two sun positions give two shadow lengths from the same height.

Definition

A vertical tower of height $h$ with the sun at elevation $\theta$ casts a horizontal shadow of length $s$ , where the sun's elevation IS the angle in the height-over-shadow triangle:

\tan\theta = \frac{h}{s} \;\Longrightarrow\; s = \frac{h}{\tan\theta} = h\cot\theta.

A lower sun (smaller $\theta$ ) gives a longer shadow.
When the elevation changes from $\theta_1$ to $\theta_2$ , the shadow changes by $\;s_2 - s_1 = h(\cot\theta_2 - \cot\theta_1)$ — set this equal to the given lengthening to find $h$ or the angle.

Shadow of a vertical object

s = h\cot\theta, \qquad \Delta s = h(\cot\theta_2 - \cot\theta_1)

Worked example

A tower of height

20\sqrt{3}

m casts a shadow that grows by

x

m when the sun's elevation drops from

60^\circ

30^\circ

. Find

x

Shadow at $60^\circ$ : $s_1 = h\cot 60^\circ = 20\sqrt{3}\cdot\dfrac{1}{\sqrt{3}} = 20$ m.
Shadow at $30^\circ$ : $s_2 = h\cot 30^\circ = 20\sqrt{3}\cdot\sqrt{3} = 60$ m.
Lengthening: $x = s_2 - s_1 = 60 - 20 = 40$ m.

Answer:

x = 40

Practice this conceptself-check

Try it yourself

6

m flagstaff on top of a tower casts a ground shadow of

2\sqrt{3}

m (the flagstaff's own shadow). What angle does the sun make with the ground?

From the bank · past-year question

Example 1Height & DistanceHARD

The shadow of a tower is found to be

x

metre longer, when the angle of elevation of the sun changes from 60° to 45°. If the height of the tower is

5(3+\sqrt{3})

m, then what is

x

equal to?

[Q29 · Apr · 2021]

Lower sun, longer shadow

Because

s = h\cot\theta

and cotangent decreases as the angle grows, the shadow is LONGER when the elevation is SMALLER. If your lengthening

s_2 - s_1

comes out negative, you have the two angles swapped.

Drill 1 more on shadows and the sun's elevation

Concept 2 of 5

Finding the New Sun Angle from a Shadow Change

Intuition

Sometimes the height is tied to the shadow change itself, and the question asks which range the new elevation falls in. You compute the new tangent, then compare it against the standard tangents 1/√3, 1, √3 to box the angle between two known values.

Definition

When the height is given in terms of the shadow lengthening $x$ (e.g. $h = \sqrt{3}\,x$ ) and the sun drops from $\theta_1$ to $\theta$ :

First shadow: $s_1 = h\cot\theta_1$ . New shadow: $s = s_1 + x$ .
New tangent: $\tan\theta = \dfrac{h}{s_1 + x}$ .
Bracket the angle by comparing $\tan\theta$ with the reference values $\tan 30^\circ = \tfrac{1}{\sqrt{3}} \approx 0.577$ , $\tan 45^\circ = 1$ , $\tan 60^\circ = \sqrt{3} \approx 1.732$ .

New tangent, then bracket

\tan\theta = \frac{h}{s_1 + x}

Worked example

A tower's shadow grows by

x

when the sun drops from

60^\circ

\theta

. If the height is

\sqrt{3}\,x

, in which range does

\theta

lie?

At $60^\circ$ : $s_1 = h\cot 60^\circ = \dfrac{\sqrt{3}x}{\sqrt{3}} = x$ .
New shadow: $s = s_1 + x = 2x$ , so $\tan\theta = \dfrac{h}{2x} = \dfrac{\sqrt{3}x}{2x} = \dfrac{\sqrt{3}}{2} \approx 0.866$ .
Since $\tan 30^\circ \approx 0.577 < 0.866 < 1 = \tan 45^\circ$ , we get $30^\circ < \theta < 45^\circ$ .

Answer:

30^\circ < \theta < 45^\circ

(about

40.9^\circ

From the bank · past-year question

Example 2Height & DistanceHARD

The shadow of a tower becomes

x

metre longer, when the angle of elevation of sun changes from

60°

\theta

. If the height of the tower is

\sqrt{3}x

metre, then which one of the following is correct?

[Q35 · Apr · 2022]

The new shadow is old + increase, not just the increase

The lengthening

x

adds to the original shadow

s_1

; the new shadow is

s_1 + x

. Dividing the height by

x

alone (instead of

s_1 + x

) over-estimates the tangent and lands you in the wrong angle band.

Concept 3 of 5

Leaning Towers — Separating Height from Lean

Intuition

A tower tilted off the vertical has its top NOT over its foot — there is a horizontal lean as well as a vertical height. One elevation reading can't separate the two unknowns, so you take two readings from points on opposite sides and solve the pair.

Definition

A leaning tower has vertical height $h$ and its top is shifted a horizontal distance $\delta$ from the foot. Reading the top's elevation from two ground points $P$ (distance $p$ ) and $Q$ (distance $q$ ) on the same line:

$\tan(\text{angle at }P) = \dfrac{h}{p - \delta}$ , $\quad\tan(\text{angle at }Q) = \dfrac{h}{q - \delta}$ .
Two equations, two unknowns $(h, \delta)$ — solve them together. With the classic $15^\circ$ and $75^\circ$ pair, $\tan 15^\circ = 2-\sqrt{3}$ and $\tan 75^\circ = 2+\sqrt{3}$ give a clean answer like $h = \dfrac{x-y}{2\sqrt{3}}$ .
The tower's inclination $\theta$ to the horizontal satisfies $\cot\theta = \dfrac{\delta}{h}$ , and its actual length along the slant is $\dfrac{h}{\sin\theta}$ .

Two readings on a leaning tower

\tan\alpha = \frac{h}{p-\delta}, \qquad \tan\beta = \frac{h}{q-\delta}

Worked example

A tower leans so its top is shifted

\delta

horizontally from its foot, height

h

. From

P

(distance

x

) the top's elevation is

15^\circ

; from

Q

(distance

y

, nearer, same side) it is

75^\circ

. Find

h

From $P$ : $\tan 15^\circ = \dfrac{h}{x-\delta} = 2-\sqrt{3}$ . From $Q$ : $\tan 75^\circ = \dfrac{h}{y-\delta} = 2+\sqrt{3}$ .
So $x - \delta = \dfrac{h}{2-\sqrt{3}} = h(2+\sqrt{3})$ and $y - \delta = \dfrac{h}{2+\sqrt{3}} = h(2-\sqrt{3})$ .
Subtract: $(x-\delta)-(y-\delta) = x - y = h[(2+\sqrt{3})-(2-\sqrt{3})] = 2\sqrt{3}\,h\Rightarrow h = \dfrac{x-y}{2\sqrt{3}}$ .

Answer:

h = \dfrac{x-y}{2\sqrt{3}}

From the bank · past-year question

Example 3Height & DistanceHARD

Consider the following for the items that follow: There are two points P and Q due south of a leaning tower, which leans towards north. P is at a distance x and Q is at a distance y from the foot of the tower (x > y). The angles of elevation of the top of the tower from P and Q are 15° and 75° respectively.

At what height is the top of the tower above the ground level?

[Q43 · Sep · 2022]

A leaning tower has two unknowns

Vertical height

h

AND horizontal lean

\delta

are both unknown, so a single elevation reading is not enough — you must use both observation points. Treating the leaning tower like a vertical one (assuming

\delta = 0

) is the trap the whole set is built around.

Drill 2 more on leaning towers — separating height from lean

Concept 4 of 5

Chord Length of a Circle

Intuition

A chord that subtends an angle at the centre of a circle splits into two right triangles by the radius that bisects it. Half the chord is the opposite side of half the central angle, so the whole chord is twice the radius times the sine of the half-angle.

Definition

A chord of a circle of radius $r$ subtending a central angle $\theta$ has length

\text{chord} = 2r\sin\frac{\theta}{2}.

The radius to the chord's midpoint is perpendicular and bisects both the chord and the angle, giving the right triangle with half-chord

= r\sin(\theta/2)

. For the NDA's odd angles use the half-angle value, e.g.

\sin 22.5^\circ = \dfrac{1}{2}\sqrt{2-\sqrt{2}}

Chord subtending central angle θ

\text{chord} = 2r\sin\frac{\theta}{2}

Worked example

Find the length of a chord of a unit circle (

r = 1

) that subtends

45^\circ

at the centre.

Chord $= 2r\sin(\theta/2) = 2\sin 22.5^\circ$ .
Half-angle: $\sin 22.5^\circ = \sqrt{\dfrac{1-\cos 45^\circ}{2}} = \dfrac{1}{2}\sqrt{2-\sqrt{2}}$ .
Chord $= 2\cdot\dfrac{1}{2}\sqrt{2-\sqrt{2}} = \sqrt{2-\sqrt{2}}$ .

Answer:

\sqrt{2-\sqrt{2}}

units.

From the bank · past-year question

Example 4Height & DistanceMODERATE

What is the length of the chord of a unit circle which subtends at the centre of the circle an angle of 45°?

[Q40 · Apr · 2026]

Half the angle, not the whole angle

The chord formula uses

\sin(\theta/2)

, because the perpendicular radius bisects the central angle. Writing

2r\sin\theta

is the standard slip — and it gives a length that can wrongly exceed the diameter.

Concept 5 of 5

Arc Length and the Equilateral-Chord Clue

Intuition

Arc length is just radius times the central angle in radians. The hidden step is finding the angle: when a chord equals the radius, the chord and the two radii form an equilateral triangle, so the central angle is 60°.

Definition

For a circle of radius $r$ , an arc subtending a central angle $\theta$ (in radians) has length

\text{arc} = r\theta.

A useful recognition: a chord of length equal to the radius makes an equilateral triangle with the two radii, so it subtends $60^\circ = \dfrac{\pi}{3}$ at the centre.
Convert any degree angle to radians before multiplying; use $\pi \approx \dfrac{22}{7}$ when the options are fractions.

Arc length

\text{arc} = r\theta \quad (\theta \text{ in radians})

Worked example

A circle has diameter

44

cm and a chord of length

22

cm. Find the length of the minor arc cut off by the chord. (

\pi \approx \tfrac{22}{7}

Radius $r = 22$ cm, and the chord is $22$ cm $= r$ , so the chord and two radii form an equilateral triangle: central angle $= 60^\circ = \dfrac{\pi}{3}$ .
Arc $= r\theta = 22\cdot\dfrac{\pi}{3} = \dfrac{22\pi}{3}$ .
With $\pi \approx \dfrac{22}{7}$ : arc $= \dfrac{22\cdot 22}{3\cdot 7} = \dfrac{484}{21}$ cm.

Answer:

\dfrac{484}{21}

\approx 23.0

cm.

From the bank · past-year question

Example 5Height & DistanceMODERATE

In a circle of diameter 44 cm, the length of a chord is 22 cm. What is the length of minor arc of the chord?

[Q97 · Apr · 2019]

Angle must be in radians for r·θ

Arc length

= r\theta

only when

\theta

is in radians. Plugging in

60

(degrees) instead of

\pi/3

gives an answer nearly

57\times

too big. Convert first.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Shadows and the Sun's Elevation
Shadow of a vertical object
$s = h\cot\theta, \qquad \Delta s = h(\cot\theta_2 - \cot\theta_1)$
Finding the New Sun Angle from a Shadow Change
New tangent, then bracket
$\tan\theta = \frac{h}{s_1 + x}$
Leaning Towers — Separating Height from Lean
Two readings on a leaning tower
$\tan\alpha = \frac{h}{p-\delta}, \qquad \tan\beta = \frac{h}{q-\delta}$
Chord Length of a Circle
Chord subtending central angle θ
$\text{chord} = 2r\sin\frac{\theta}{2}$
Arc Length and the Equilateral-Chord Clue
Arc length
$\text{arc} = r\theta \quad (\theta \text{ in radians})$

Watch out for (5)

Lower sun, longer shadow
→ Shadows and the Sun's Elevation
The new shadow is old + increase, not just the increase
→ Finding the New Sun Angle from a Shadow Change
A leaning tower has two unknowns
→ Leaning Towers — Separating Height from Lean
Half the angle, not the whole angle
→ Chord Length of a Circle
Angle must be in radians for r·θ
→ Arc Length and the Equilateral-Chord Clue

Mastery check — 3 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Height & DistanceHARD

A flag-staff of 6 m on top of a tower throws shadow of

2\sqrt{3}

m along the ground. What is the angle the sun makes with the ground?

[Q43 · Apr · 2018]

Example 2Height & DistanceHARD

\theta

is the inclination of the tower to the horizontal, then what is

\cot\theta

equal to?

[Q44 · Sep · 2022]

Example 3Height & DistanceHARD

What is the length of the tower?

[Q45 · Sep · 2022]

Drill every past-year question on this subtopic

8 questions from the bank — paginated, with cart and Word-export support.

Drill the 8 questions