Heights and Distances

high importance~2 Q in Tier 117 formulas⚡ 10 shortcuts5 subtopics

Overview Learn (5)Formulas (17)Examples (5)Practice (67)

Applied trigonometry: angles of elevation and depression, the 30-45-60 rules, two-angle configurations, moving observers and compound figures. Every question reduces to one right triangle and one tangent — learn the three standard-angle multipliers and the cot-difference formula and this becomes near-guaranteed marks.

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Learn — 5 subtopics

One page per subtopic: detailed notes, every question type, formulas, tricks and practice sets.

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Formula sheet

Every formula on one printable page, grouped by subtopic.

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Solved examples

5 exam-level questions worked step by step.

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Practice & test

67 questions — untimed practice or a timed test with analysis.

Track record in the exam

avg 1.0 Q / shift2024: 1 Q2025: 1 Q

Questions per shift in recent SSC CGL papers.

Test difficulty mix (67 questions)

17 easy33 medium17 hard

Question patterns exams keep repeating

Taken from previous-year papers. If a pattern is marked "very common", expect to see it in your exam.

Single angle, one unknown

very common

Spot it:

One angle (30°, 45° or 60°) with either the height or the distance given.

How to solve: Standard-angle rule: $d = h$ at 45°, $d = \sqrt3 h$ at 30°, $d = \frac{h}{\sqrt3}$ at 60°; keep the radical until the option match.

Example: The angle of elevation of the top of a 50 m tower from a point on the ground is $60^\circ$ . The distance of the point from the base is:

$d = \frac{50}{\sqrt3} = \frac{50\sqrt3}{3}$ m.

Learn this in “Angles of elevation and depression” →

Angle of depression from a height

very common

Spot it:

Lighthouse/cliff/aeroplane looking down at ships, cars or buildings.

How to solve: Depression at the eye = elevation at the object; then $d = h\cot\theta$ . Two objects → subtract (same side) or add (opposite sides) the two distances.

Example: From a 100 m cliff the depression of a boat is $30^\circ$ . Its distance from the cliff base is:

$d = 100\tan60^\circ = 100\sqrt3$ m.

Learn this in “Angles of elevation and depression” →

Two angles from two points (walking or opposite sides)

very common

Spot it:

Elevation changes from α to β after walking $w$ m, or two observers $D$ apart on opposite sides.

How to solve: $h = \frac{w}{\cot\alpha - \cot\beta}$ (same side) or $\frac{D}{\cot\alpha + \cot\beta}$ (opposite); 30–60 pairs give $\frac{w\sqrt3}{2}$ and $\frac{D\sqrt3}{4}$ .

Example: Walking 40 m towards a tower changes the elevation from $30^\circ$ to $60^\circ$ . The height is:

$h = \frac{40\sqrt3}{2} = 20\sqrt3$ m.

Learn this in “Two observation points (two angles)” →

Speed-time two-angle problem

common

Spot it:

A car/boat moves at uniform speed; the depression or elevation changes after a stated time.

How to solve: Distance = speed × time $= h(\cot\theta_{\text{far}} - \cot\theta_{\text{near}})$ ; solve for $h$ , $v$ or $t$ . Remaining distance to the base is $h\cot\theta_{\text{near}}$ .

Example: At 6 m/s a car's depression changes from $30^\circ$ to $60^\circ$ in 6 s. The tower's height is:

$36 = \frac{2h}{\sqrt3} \Rightarrow h = 18\sqrt3$ m.

Learn this in “Moving observers: speed and time” →

Compound structure (statue/pedestal, tower/building)

common

Spot it:

A statue on a pedestal, tower on a building, or flagstaff on a tower; two elevations from one point.

How to solve: Split at the change point: upper piece $= d(\tan\beta - \tan\alpha)$ ; a 45° lower angle fixes $d$ = lower height. Roof observations: add the building back.

Example: A statue on a 30 m pedestal is seen at $45^\circ$ (pedestal top) and $60^\circ$ (statue top). The statue's height is:

$30\tan60^\circ - 30 = 30(\sqrt3-1)$ m.

Learn this in “Compound figures: buildings, pedestals, broken objects” →

Shadow problems

common

Spot it:

Shadow length related to height (equal, √3 times, or a ratio), possibly two objects at once.

How to solve: Shadow = horizontal distance: $\tan(\text{elevation}) = \frac{h}{\text{shadow}}$ . Equal shadows ⇒ equal heights; two objects at one moment share the sun's angle — use ratios.

Example: A stick 1.5 m casts a 2.5 m shadow while a tower casts 75 m. The tower's height is:

$75 \times \frac{1.5}{2.5} = 45$ m.

Learn this in “Angles of elevation and depression” →

Broken tree / pole

occasional

Spot it:

A tree breaks and the top touches the ground at a stated distance and ground angle.

How to solve: Standing $= b\tan\theta$ ; broken (hypotenuse) $= \frac{b}{\cos\theta}$ ; original height $= b\tan\theta + \frac{b}{\cos\theta}$ .

Example: A tree breaks; the top touches ground 15 m away at $30^\circ$ . The original height is:

$5\sqrt3 + 10\sqrt3 = 15\sqrt3$ m.

Learn this in “Compound figures: buildings, pedestals, broken objects” →

Hypotenuse given (ladder, thread, wire)

occasional

Spot it:

The slant distance is given with the angle; a height or ground distance asked.

How to solve: $h = L\sin\theta$ , $d = L\cos\theta$ — sine/cosine, never tangent, when the hypotenuse is the given.

Example: A 60 m cable makes $60^\circ$ with the ground. The pole height is:

$60\sin60^\circ = 30\sqrt3$ m.

Learn this in “Angles of elevation and depression” →

Your next step

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