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Heights and Distances

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high importance~2 Q in Tier 117 formulas⚡ 10 shortcuts5 subtopics
OverviewLearn (5)Formulas (17)Examples (5)Practice (67)
Subtopic 3 of 5·← Standard angles: 30°, 45°, 60°·Moving observers: speed and time →

Two observation points (two angles)

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The highest-yield pattern: two angles from two points on the same line.

  • Same side (walk toward the object, angle grows from α\alpha to β\beta, β>α\beta>\alpha): the distance walked equals h(cot⁡α−cot⁡β)h(\cot\alpha-\cot\beta), so h=dcot⁡α−cot⁡βh=\frac{d}{\cot\alpha-\cot\beta}
  • Opposite sides of the object, points dd apart: h=dcot⁡α+cot⁡βh=\frac{d}{\cot\alpha+\cot\beta}
  • From a height (two ships seen from a lighthouse): the gap between them is h(cot⁡α−cot⁡β)h(\cot\alpha-\cot\beta) for angles α<β\alpha<\beta measured to the nearer ship... i.e. gap =dfar−dnear=d_{\text{far}}-d_{\text{near}}.

With 30°/60° pairs remember cot⁡30∘=3\cot30^\circ=\sqrt3, cot⁡60∘=13\cot60^\circ=\frac{1}{\sqrt3}, so cot⁡30−cot⁡60=23\cot30-\cot60=\frac{2}{\sqrt3} and h=d32h=\frac{d\sqrt3}{2} — the answer is 32\frac{\sqrt3}{2} of the walked distance.

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

Detailed notes

The two cotangent distances

Stand d1d_1 from a tower of height hh: the elevation is α\alpha, so d1=hcot⁡αd_1 = h\cot\alpha. Walk closer to d2d_2: the elevation grows to β\beta, and d2=hcot⁡βd_2 = h\cot\beta. Every two-angle question is these two equations plus one piece of information linking d1d_1 and d2d_2.

Same side: the difference formula

Walking a distance ww toward the object: w=d1−d2=h(cot⁡α−cot⁡β)w = d_1 - d_2 = h(\cot\alpha - \cot\beta), so h=wcot⁡α−cot⁡βh = \frac{w}{\cot\alpha - \cot\beta} The exam's favourite pair is 30∘30^\circ–60∘60^\circ: cot⁡30∘−cot⁡60∘=3−13=23\cot30^\circ - \cot60^\circ = \sqrt3 - \frac{1}{\sqrt3} = \frac{2}{\sqrt3}, giving h=w32h = \frac{w\sqrt3}{2} — the height is about 0.870.87 of the walked distance. The 30–45 pair gives h=w3−1=w(3+1)2h = \frac{w}{\sqrt3 - 1} = \frac{w(\sqrt3+1)}{2}.

Opposite sides: the sum formula

Two observers on opposite sides, DD apart: their distances add to DD, so h=Dcot⁡α+cot⁡βh = \frac{D}{\cot\alpha + \cot\beta} For the 30–60 pair: cot⁡30+cot⁡60=3+13=43\cot30 + \cot60 = \sqrt3 + \frac{1}{\sqrt3} = \frac{4}{\sqrt3}, so h=D34h = \frac{D\sqrt3}{4}. The sign inside the denominator is the whole difference between the two configurations — subtract for same side, add for opposite sides.

Gap problems from a height

From a lighthouse hh high, two ships are seen at depressions α\alpha (far) and β\beta (near). Their distances from the base are hcot⁡αh\cot\alpha and hcot⁡βh\cot\beta; the gap between them is h(cot⁡α−cot⁡β)h(\cot\alpha - \cot\beta) — the same machinery, worked top-down. If the ships are on opposite sides, the two distances add instead.

Two towers from a midpoint

Viewed from the midpoint of the line joining two bases, the elevations of two tower tops are α\alpha and β\beta. Half the separation fixes each base distance mm; then h1=mtan⁡αh_1 = m\tan\alpha, h2=mtan⁡βh_2 = m\tan\beta, and the difference is m(tan⁡β−tan⁡α)m(\tan\beta - \tan\alpha). Heights come from tan here (each tower owns its own triangle), unlike the single-tower case which is cleanest in cot.

Which formula applies? (a 5-second test)

One object, two observation points: ask "are the two points on the same side of the object's base?" Same side (you walked toward it) → difference of cots. Opposite sides → sum of cots. One observer, two objects: each object gets its own cot distance from the same base point; the asked gap is their difference (same side) or sum (opposite sides). The words "walking towards", "approaching" and "on the same side" all signal subtraction; "on opposite sides", "on either side of it" signal addition.

Quick revision

  • d=hcot⁡θd = h\cot\theta at any observation; two observations → two equations.
  • Same side (walk ww): h=wcot⁡α−cot⁡βh = \frac{w}{\cot\alpha - \cot\beta}; nearer point has the LARGER angle.
  • Opposite sides (apart DD): h=Dcot⁡α+cot⁡βh = \frac{D}{\cot\alpha + \cot\beta} — add, don't subtract.
  • 30–60 pair: cot⁡30−cot⁡60=23\cot30 - \cot60 = \frac{2}{\sqrt3}, cot⁡30+cot⁡60=43\cot30 + \cot60 = \frac{4}{\sqrt3}.
  • Gap from a height: h(cot⁡α−cot⁡β)h(\cot\alpha - \cot\beta), far minus near.
  • Two towers from a midpoint: m(tan⁡β−tan⁡α)m(\tan\beta - \tan\alpha), heights in tan.

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: Walking toward: two angles, same sidevery common3 practice Q
How to spot it:

'On walking x m towards the tower the elevation changes from α to β' — find the height (or the initial distance).

$h = \frac{w}{\cot\alpha - \cot\beta}$; the 30–60 pair gives $h = \frac{w\sqrt3}{2}$ instantly.
  1. Write both distances as cot products of h.
  2. Their difference is the walked distance — one linear equation.
  3. For the distance asked instead, get h first, then hcot⁡βh\cot\beta (the near distance).

Why: subtracting two cot expressions collapses the unknown near-distance — the reason this formula exists.

Example: On walking 4040 m towards a tower the elevation of its top changes from 30∘30^\circ to 60∘60^\circ. The height of the tower is:

hcot⁡30∘−hcot⁡60∘=40⇒h⋅23=40⇒h=203h\cot30^\circ - h\cot60^\circ = 40 \Rightarrow h \cdot \frac{2}{\sqrt3} = 40 \Rightarrow h = 20\sqrt3 m.

Type 2: Observers on opposite sidesvery common3 practice Q
How to spot it:

Two people/points on opposite sides of the tower, separation given, elevations differ — height asked.

$h = \frac{D}{\cot\alpha + \cot\beta}$; 30–60 pair → $h = \frac{D\sqrt3}{4}$.
  1. The base distances ADD to the given separation.
  2. One equation, one unknown h.
  3. If the FOOT positions differ in ground level, that is a different (compound) question.

Why: "opposite sides" flips the subtraction to addition — the classic trap is reusing the same-side formula.

Example: Two men on opposite sides of a tower, 8080 m apart, observe its top at 30∘30^\circ and 60∘60^\circ. The tower's height is:

h=803+13=8034=203h = \frac{80}{\sqrt3 + \frac{1}{\sqrt3}} = \frac{80\sqrt3}{4} = 20\sqrt3 m.

Type 3: Two objects seen from a height (gap)common2 practice Q
How to spot it:

From a cliff/lighthouse, two ships/cars on the SAME side at two depressions; the gap between them asked.

Gap $= h(\cot\alpha_{\text{far}} - \cot\beta_{\text{near}})$; opposite sides → $h(\cot\alpha + \cot\beta)$.
  1. Convert each depression to a base distance hcot⁡θh\cot\theta.
  2. Same side: subtract (far − near). Opposite sides: add.
  3. If the gap is given and h asked, invert the same relation.

Why: each depression independently gives a distance; the gap is just their difference on the same baseline.

Example: From a 6060 m lighthouse the depressions of two ships on the same side are 30∘30^\circ and 60∘60^\circ. The distance between the ships is:

60cot⁡30∘−60cot⁡60∘=603−203=40360\cot30^\circ - 60\cot60^\circ = 60\sqrt3 - 20\sqrt3 = 40\sqrt3 m.

Type 4: Two towers from a midpoint / ratio of heightsoccasional2 practice Q
How to spot it:

Elevations of two tower tops from the midpoint of their bases (or a point with known base ratio); difference or ratio of heights asked.

Each height = (base distance) × tan(its angle); from the midpoint, distance is the same $m$ for both: difference $= m(\tan\beta - \tan\alpha)$.
  1. Fix the shared base distance m.
  2. Height of each tower = m·tan(its elevation).
  3. Difference or ratio follows; heights are 1:3 when the angles are 30° and 60°.

Why: two independent triangles share one horizontal — the midpoint makes the shared leg equal.

Example: The elevations of the tops of two towers from the midpoint of the line joining their bases are 30∘30^\circ and 60∘60^\circ. The towers are 6060 m apart. The difference of their heights is:

m=30m = 30: h2−h1=30tan⁡60∘−30tan⁡30∘=303−103=203h_2 - h_1 = 30\tan60^\circ - 30\tan30^\circ = 30\sqrt3 - 10\sqrt3 = 20\sqrt3 m.

Formulas

Same side
h=dcot⁡α−cot⁡βh=\frac{d}{\cot\alpha-\cot\beta}
Opposite sides
h=dcot⁡α+cot⁡βh=\frac{d}{\cot\alpha+\cot\beta}
30-60 pair
cot⁡30∘−cot⁡60∘=23⇒h=d32\cot30^\circ-\cot60^\circ=\frac{2}{\sqrt3}\Rightarrow h=\frac{d\sqrt3}{2}
Gap from a height
gap=h(cot⁡α−cot⁡β)\text{gap}=h(\cot\alpha-\cot\beta)

Shortcut tricks

⚡ Subtract the cotangents, don't solve two triangles

Write d1=hcot⁡αd_1=h\cot\alpha, d2=hcot⁡βd_2=h\cot\beta from each position; the difference of the distances is what is given. One linear equation in hh.

Example: Walking 40 m toward a tower changes the elevation angle from 30∘30^\circ to 60∘60^\circ. Find the tower's height.

hcot⁡30∘−hcot⁡60∘=40⇒h(3−13)=40⇒h⋅23=40⇒h=203h\cot30^\circ-h\cot60^\circ=40\Rightarrow h(\sqrt3-\frac1{\sqrt3})=40\Rightarrow h\cdot\frac2{\sqrt3}=40\Rightarrow h=20\sqrt3 m.

⚡ Opposite sides means add

When the two observers are on opposite sides, the base distances add up to dd: denominators add, not subtract.

Example: Two men on opposite sides of a tower 80 m apart observe its top at 30∘30^\circ and 60∘60^\circ. Find the height.

h=803+13=8043=203h=\frac{80}{\sqrt3+\frac1{\sqrt3}}=\frac{80}{\frac{4}{\sqrt3}}=20\sqrt3 m.

Where students lose marks

  • Subtracting base distances for opposite-side observers (they add).

  • Using tan instead of cot in the difference formula: d=h(cot⁡α−cot⁡β)d=h(\cot\alpha-\cot\beta), not h(tan⁡β−tan⁡α)h(\tan\beta-\tan\alpha)... note cot⁡α−cot⁡β\cot\alpha-\cot\beta with α<β\alpha<\beta is positive.

  • Forgetting the nearer point has the LARGER angle when assigning which distance is which.

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 8 min · wrong answers go to your mistake notebook automatically.