Heights and Distances

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

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Subtopic 2 of 5·← Angles of elevation and depression·Two observation points (two angles) →

Standard angles: 30°, 45°, 60°

Almost every CGL question uses only these three angles. Convert tan to a height–distance rule and the answer is a one-liner:

$\theta$	$\tan\theta$	rule ( $h$ = height, $d$ = distance)
$45^\circ$	$1$	$d=h$
$30^\circ$	$\frac{1}{\sqrt3}$	$d=\sqrt3\,h$
$60^\circ$	$\sqrt3$	$d=\frac{h}{\sqrt3}$

So height from distance: $45^\circ\to h=d$ ; $30^\circ\to h=\frac{d}{\sqrt3}$ ; $60^\circ\to h=\sqrt3\,d$ . Take $\sqrt3\approx1.732$ , $\sqrt2\approx1.414$ only if options are decimal.

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 15 practice Q

Detailed notes

The three multipliers

Almost every exam question uses only $30^\circ$ , $45^\circ$ or $60^\circ$ . Convert each to a height–distance rule and the tangent table becomes a cheat sheet:

$\theta$	$\tan\theta$	rule ( $h$ up, $d$ across)
$30^\circ$	$\frac{1}{\sqrt3}$	$d = \sqrt3\,h$
$45^\circ$	$1$	$d = h$
$60^\circ$	$\sqrt3$	$d = \frac{h}{\sqrt3}$

Read the pair as a swap: at $30^\circ$ the distance is $\sqrt3$ times the height; at $60^\circ$ the height is $\sqrt3$ times the distance. Ratios like $\frac{h}{d}$ identify the angle from either side.

Keeping surds exact

$\sqrt3 \approx 1.732$ and $\sqrt2 \approx 1.414$ — but only reach for decimals if the options are decimal. An answer of $\frac{40}{\sqrt3}$ rationalises to $\frac{40\sqrt3}{3}$ ; the options may present either form, so carry the radical and match at the end. Every "odd" distractor is a surd in the wrong slot ( $h\sqrt3$ vs $\frac{h}{\sqrt3}$ is the single most common trap).

Angles that are not 30-45-60

Elevations like $15^\circ$ , $75^\circ$ or $22.5^\circ$ do appear — as differences of standard angles. $15^\circ = 45^\circ - 30^\circ$ , $75^\circ = 45^\circ + 30^\circ$ ; the compound expressions $\tan(45^\circ \pm \theta) = \frac{1 \pm \tan\theta}{1 \mp \tan\theta}$ finish them in one line. A question using $75^\circ$ is almost always a $45+30$ identity question in disguise.

The ladder and the hypotenuse rule

When the slant length (ladder, thread, wire) is the given, switch to sine and cosine: $h = L\sin\theta$ , $d = L\cos\theta$ . The classic check: a $45^\circ$ ladder of length $L\sqrt2$ climbs exactly $L$ — the $45^\circ$ isosceles right triangle always has hypotenuse $\sqrt2$ times a leg.

Combination moves with the same angle twice

Two objects at the same standard angle from two positions of known separation (e.g. two towers seen from a point between them at $30^°$ each, separation $D$ ): the heights are $\frac{D}{2\sqrt3}$ each — equal heights, equal angles. When the angles differ, each height uses its own tangent against the half-separation. A chain of two standard-angle steps (walk 10 m at 30°, then 10 m more at 45°) is just two single-angle equations written in sequence and subtracted; nothing new is needed beyond bookkeeping.

Quick revision

$30^\circ$ : $d = \sqrt3 h$ ; $45^\circ$ : $d = h$ ; $60^\circ$ : $h = \sqrt3 d$ .
$\frac{h}{d} = \frac{1}{\sqrt3}$ → 30°, $= 1$ → 45°, $= \sqrt3$ → 60°.
Shadow $= \sqrt3 h$ at 30°; shadow $= h$ at 45°; shadow $= \frac{h}{\sqrt3}$ at 60°.
Hypotenuse given: $L\sin\theta$ up, $L\cos\theta$ across.
$15^\circ$ and $75^\circ$ are $45 \mp 30$ identities in disguise.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Height from distance (or reverse) at a standard anglevery common4 practice Q

How to spot it:

One standard angle, height or distance given, the other asked — often dressed as a tower, statue, pole or hill.

$30^\circ \to d = \sqrt3 h$; $45^\circ \to d = h$; $60^\circ \to d = \frac{h}{\sqrt3}$. Keep surds exact and match the option form at the end.

Write the rule for the stated angle before touching numbers.
Substitute; rationalise only if the options demand it.
Check the size: 60° puts the point CLOSER than the height; 30° puts it FARTHER.

Why: the whole question is one table lookup — the difficulty is only in the surd bookkeeping.

Example: The elevation of the top of a tower from a point $30\sqrt3$ m from its base is $30^\circ$ . The tower's height is:

$h = \frac{30\sqrt3}{\sqrt3} = 30$ m.

Type 2: Reading the angle from a ratiocommon2 practice Q

How to spot it:

'The shadow is √3 times the height' or 'height equals distance' — the angle itself is the answer.

$\tan\theta = \frac{h}{d}$: shadow $\sqrt3 h$ → $30^\circ$; equal → $45^\circ$; shadow $= \frac{h}{\sqrt3}$ → $60^\circ$.

Write tan as height over distance (shadow IS the distance).
Reduce the ratio to a table value.
Quote the angle — no further computation exists.

Why: the inverse direction of the standard rules; the same three numbers appear in reverse.

Example: When the sun's elevation is such that a pole's shadow equals the pole's height, the elevation is:

$\tan\theta = \frac{h}{h} = 1 \Rightarrow \theta = 45^\circ$ .

Type 3: Ladder / wire sliding geometryoccasional2 practice Q

How to spot it:

A ladder leaning at a standard angle; the height reached or the foot distance (or the length itself) asked.

$h = L\sin\theta$, $d = L\cos\theta$. $45^\circ$ → $h = d = \frac{L}{\sqrt2}$; $30^\circ$ → $h = \frac{L}{2}$; $60^\circ$ → $h = \frac{L\sqrt3}{2}$.

The ladder is the hypotenuse — sin and cos only.
At 30° the vertical is exactly half the ladder (sin30 = ½).
Sanity-check with the isosceles case: 45° puts the foot at $L/\sqrt2$ .

Why: 30-60-90 triangles have sides $1 : \sqrt3 : 2$ — half the ladder is the shortest side.

Example: A ladder $20$ m long rests at $30^\circ$ against a wall. How high up the wall does it reach?

$h = 20\sin30^\circ = 10$ m.

Type 4: 45+30 identity angles (15°, 75°)occasional2 practice Q

How to spot it:

Elevation stated as 15° or 75°, or a tan(45°±θ) expression appears inside the question.

$\tan(45^\circ \pm \theta) = \frac{1 \pm \tan\theta}{1 \mp \tan\theta}$; with $\theta = 30^\circ$: $\tan75^\circ = 2+\sqrt3$, $\tan15^\circ = 2-\sqrt3$.

Split the odd angle as 45 ± 30.
Apply the compound formula; memorise the two finished values.
Expect a factored surd answer $(2 \pm \sqrt3)$ — do not decimalise.

Why: 15°/75° triangles are the standard "hard" elevations; the formula turns them into one substitution.

Example: The value of $\tan 15^\circ \tan 75^\circ$ is:

$(2-\sqrt3)(2+\sqrt3) = 4 - 3 = 1$ (they are complementary).

Formulas

45 degrees

\theta=45^\circ\Rightarrow d=h

30 degrees

\theta=30^\circ\Rightarrow d=\sqrt3\,h\ \Leftrightarrow\ h=\frac{d}{\sqrt3}

60 degrees

\theta=60^\circ\Rightarrow d=\frac{h}{\sqrt3}\ \Leftrightarrow\ h=\sqrt3\,d

Ladder (hypotenuse given)

h=L\sin\theta,\quad d=L\cos\theta

Shortcut tricks

⚡ Match the angle to the multiplier

45° → same; 30° → distance is $\sqrt3$ times the height; 60° → height is $\sqrt3$ times the distance. Write the multiplier before touching numbers.

Example: A tower subtends $60^\circ$ at a point 20 m from its base. Find its height.

$h=\sqrt3\times20=20\sqrt3$ m.

⚡ Ladder questions are sine/cosine, not tangent

The ladder itself is the hypotenuse: height $=L\sin\theta$ , foot distance $=L\cos\theta$ . A $45^\circ$ ladder of length $L\sqrt2$ climbs exactly $L$ .

Example: A ladder of length $10\sqrt2$ m leans at $45^\circ$ . How high up the wall does it reach?

$10\sqrt2\times\frac{1}{\sqrt2}=10$ m.

Where students lose marks

Using the $60^\circ$ rule for $30^\circ$ (they are reciprocals: $d=\sqrt3h$ vs $d=h/\sqrt3$ ).
Approximating $\sqrt3$ too early and landing between two options — keep the radical until the end.
Ladder: applying $\tan\theta=h/d$ with the ladder length mistaken for $d$ .

Practice sets — 15 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 10 questions

Suggested time 5 min · wrong answers go to your mistake notebook automatically.