Heights and Distances

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high importance~2 Q in Tier 120 formulas⚡ 11 shortcuts5 subtopics

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Two observation points (two angles)

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⏱ 5 min read🧩 5 question types🎯 15 practice Q

The idea in one minute

One angle gives only a ratio: two sides, one equation. Two angles from two places give the height itself.

The trick is simple. Write $h\cot\alpha$ for one distance and $h\cot\beta$ for the other. The walked gap between the two viewpoints is the difference (same side) or the sum (opposite sides) of these.

So every two-angle question is one equation with one unknown: $h$ .

Why two angles are enough

A single angle only tells you a ratio. Add a second viewpoint and the ground between them, and the height falls out.

Set the tower's height as $h$ . From point P the angle is $\alpha$ , from point Q it is $\beta$ . Then:

Distance of P from the foot $= h\cot\alpha$ .
Distance of Q from the foot $= h\cot\beta$ .

Everything else is adding or subtracting these two lengths.

Rule: Same side of the tower: the gap is a difference. Opposite sides: the gap is a sum.

Walking towards the tower

The most common pattern. A man sees the top at $30^\circ$ , walks some distance towards it, and sees $60^\circ$ .

h(\cot\alpha - \cot\beta) = d \qquad\Rightarrow\qquad h = \frac{d}{\cot\alpha - \cot\beta}

The walked distance is 40 m: $h\left(\sqrt{3} - \dfrac{1}{\sqrt{3}}\right) = h \times \dfrac{2}{\sqrt{3}} = 40$ , so $h = 20\sqrt{3}$ m.

Tip: For the $30^\circ$ and $60^\circ$ pair, $\cot 30^\circ - \cot 60^\circ = \dfrac{2}{\sqrt{3}}$ . So $h = d \times \dfrac{\sqrt{3}}{2}$ : with $d = 40$ , $h = 20\sqrt{3}$ m in one step.

Watching from opposite sides

Two men stand on opposite banks of a river, or the two angles at the tower top are measured from two sides.

h = \frac{d}{\cot\alpha + \cot\beta}

Two points are 80 m apart on opposite sides of a tower, with angles $30^\circ$ and $60^\circ$ : $h\left(\sqrt{3} + \dfrac{1}{\sqrt{3}}\right) = h \times \dfrac{4}{\sqrt{3}} = 80$ , so $h = 20\sqrt{3}$ m.

Watch: If a question says "the angles of elevation of the top of an unfinished tower", the taller part still counts: draw the full height in both triangles.

Two objects seen from a height

From the top of a lighthouse $h$ metres high, two boats lie in the same line, at depressions $\alpha$ and $\beta$ .

Each boat's distance from the foot is $h\cot\alpha$ and $h\cot\beta$ . The distance between the boats is:

\text{gap} = h(\cot\alpha - \cot\beta)

From a 60 m lighthouse the depressions are $30^\circ$ and $60^\circ$ : gap $= 60 \times \dfrac{2}{\sqrt{3}} = 40\sqrt{3}$ m.

Two towers from one point

Two towers stand on opposite sides of a road. From a point on the road, their tops are at $\alpha$ and $\beta$ .

Each tower makes its own triangle sharing the same ground distance $d$ : heights are $d\tan\alpha$ and $d\tan\beta$ .

From the midpoint of a 60 m road, the angles are $60^\circ$ and $30^\circ$ . Each tower is 30 m away: heights $= 30\sqrt{3}$ m and $10\sqrt{3}$ m. Difference $= 20\sqrt{3}$ m.

Watching from two floors

A tower is seen from the foot of a building and again from its roof, $b$ metres higher.

From the foot the angle is $\beta$ (larger). From the roof it is $\alpha$ (smaller), because the roof already covers part of the height. The two equations are $H = d\tan\beta$ and $H - b = d\tan\alpha$ . Subtract:

d = \frac{b}{\tan\beta - \tan\alpha}, \qquad H = d\tan\beta

Building 10 m, roof $30^\circ$ , foot $60^\circ$ : $d = \dfrac{10}{\sqrt{3} - 1/\sqrt{3}} = 5\sqrt{3}$ m and $H = 5\sqrt{3} \times \sqrt{3} = 15$ m.

Watch: From the higher point the angle is smaller, because the tower top is closer to eye level.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common3 practice Q

Walking towards the tower

How to spot it:

The angle of elevation increases as the observer walks a given distance towards the tower.

h = \frac{d}{\cot\alpha - \cot\beta}

Method

Name the two angles: far first, near second.
Write both distances from the foot: $h\cot\alpha$ and $h\cot\beta$ .
Their difference equals the distance walked.
Solve for $h$ .

Why it works:

The same height sits inside two triangles, so the walked ground equals the difference of the two cotangent distances.

Try this

The angle of elevation of the top of a tower from a point is $30^\circ$ . On walking 40 m towards the tower, the angle becomes $60^\circ$ . The height of the tower is:

Show solution

Far distance $= h\cot 30^\circ = h\sqrt{3}$ .
Near distance $= h\cot 60^\circ = \dfrac{h}{\sqrt{3}}$ .
Difference: $\dfrac{2h}{\sqrt{3}} = 40$ , so $h = 20\sqrt{3}$ m.

Answer

$20\sqrt{3}$ m

Type 2common3 practice Q

Observers on opposite sides

How to spot it:

Two people (or banks of a river) on opposite sides of the tower with different angles.

h = \frac{d}{\cot\alpha + \cot\beta}

Method

Write the two distances from the foot as $h\cot\alpha$ and $h\cot\beta$ .
Their sum is the full separation $d$ .
Solve for $h$ .

Why it works:

On opposite sides the two distances add up to the ground between the observers.

Try this

Two men on opposite sides of a tower, 80 m apart, see its top at $30^\circ$ and $60^\circ$ . The height of the tower is:

Show solution

$h\cot 30^\circ + h\cot 60^\circ = 80$ .
$h\left(\sqrt{3} + \dfrac{1}{\sqrt{3}}\right) = h \times \dfrac{4}{\sqrt{3}} = 80$ .
$h = 20\sqrt{3}$ m.

Answer

$20\sqrt{3}$ m

Type 3common2 practice Q

Distance between two objects seen from a height

How to spot it:

From the top of a lighthouse or cliff, two boats or cars are seen at two angles of depression.

\text{gap} = h(\cot\alpha - \cot\beta)

Method

Convert each depression into an elevation at the object.
Distance of the farther object $= h\cot\alpha$ (smaller angle).
Subtract the nearer distance $h\cot\beta$ to get the gap.

Why it works:

Each object sits at its own cotangent distance from the foot, so the gap is their difference.

Try this

From the top of a lighthouse 60 m above the sea, the angles of depression of two boats in the same line are $30^\circ$ and $60^\circ$ . The distance between the boats is:

Show solution

Far boat: $60 \times \cot 30^\circ = 60\sqrt{3}$ m.
Near boat: $60 \times \cot 60^\circ = \dfrac{60}{\sqrt{3}} = 20\sqrt{3}$ m.
Gap $= 60\sqrt{3} - 20\sqrt{3} = 40\sqrt{3}$ m.

Answer

$40\sqrt{3}$ m

Type 4common2 practice Q

Two towers seen from one point

How to spot it:

Two towers on opposite sides of a road; from a point between them both tops are seen.

H_1 = d\tan\alpha,\qquad H_2 = d\tan\beta

Method

Split the ground distance as the question directs (midpoint means half each).
Each tower forms its own right triangle.
Compute each height; subtract if a difference is asked.

Why it works:

The triangles are independent, joined only by the shared ground distance.

Try this

Two towers stand on opposite sides of a road 60 m wide. From the midpoint of the road, their tops are seen at $60^\circ$ and $30^\circ$ . The difference of their heights is:

Show solution

Each tower is 30 m from the midpoint.
Taller tower $= 30\tan 60^\circ = 30\sqrt{3}$ m; other $= 30\tan 30^\circ = 10\sqrt{3}$ m.
Difference $= 20\sqrt{3}$ m.

Answer

$20\sqrt{3}$ m

Type 5occasional

Tower seen from the foot and the roof of a building

How to spot it:

The same tower's top is observed from ground level and from a window or roof above.

d = \frac{b}{\tan\beta - \tan\alpha}

Method

Let the building be $b$ and the tower $H$ , with ground distance $d$ .
From the foot: $\tan\beta = H/d$ . From the roof: $\tan\alpha = (H-b)/d$ .
Subtract to eliminate $H$ and find $d$ , then $H = d\tan\beta$ .

Why it works:

Raising the observer cuts the tower's apparent height by exactly the building's height.

Try this

A tower is seen from the foot of a 10 m building at $60^\circ$ , and from its roof at $30^\circ$ . The height of the tower is:

Show solution

$H - 10 = d\tan 30^\circ = \dfrac{d}{\sqrt{3}}$ and $H = d\tan 60^\circ = d\sqrt{3}$ .
$d\sqrt{3} - \dfrac{d}{\sqrt{3}} = 10 \Rightarrow \dfrac{2d}{\sqrt{3}} = 10$ .
$d = 5\sqrt{3}$ m, so $H = 5\sqrt{3} \times \sqrt{3} = 15$ m.

Answer

15 m

Formula sheet

Same side (walk towards)

h = \frac{d}{\cot\alpha - \cot\beta}

d is the distance walked; beta is the nearer, bigger angle.

Opposite sides

h = \frac{d}{\cot\alpha + \cot\beta}

d is the full distance between the two observers.

Gap between two objects from a height

\text{gap} = h(\cot\alpha - \cot\beta)

Tower seen from foot and roof of a building

d = \frac{b}{\tan\beta - \tan\alpha},\quad H = d\tan\beta

b = building height; beta from the foot, alpha from the roof.

Shortcuts that save time

⚡ Thirty-sixty fast numbers

For the 30 and 60 pair: same side $h = d\times\dfrac{\sqrt{3}}{2}$ ; opposite sides $h = d\times\dfrac{\sqrt{3}}{4}$ . Both come from $\cot 30^\circ - \cot 60^\circ = \dfrac{2}{\sqrt{3}}$ .

Example

Walking 20 m towards a tower raises the angle from $30^\circ$ to $60^\circ$ . Find the height.

Show solution

$h = 20 \times \dfrac{\sqrt{3}}{2}$ .
$h = 10\sqrt{3}$ m.

Answer

$10\sqrt{3}$ m

⚡ Write the cot pair before the numbers

Always simplify $\cot\alpha - \cot\beta$ (or the sum) first. Substituting numbers into unsimplified surds is where errors creep in.

Example

Two observers 80 m apart on opposite sides see a tower top at $45^\circ$ and $60^\circ$ . Find the height.

Show solution

$\cot 45^\circ + \cot 60^\circ = 1 + \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}+1}{\sqrt{3}}$ .
$h = \dfrac{80\sqrt{3}}{\sqrt{3}+1} = \dfrac{80\sqrt{3}(\sqrt{3}-1)}{2}$ .
$h = 40\sqrt{3}(\sqrt{3}-1) = 120 - 40\sqrt{3}$ m.

Answer

$(120-40\sqrt{3})$ m

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using a difference of cotangents for observers on opposite sides.

Opposite sides add the distances, so use $\cot\alpha+\cot\beta$ .

Mistake 02

Subtracting the angles first and then taking one cotangent.

The cotangents subtract, never the angles.

Mistake 03

Treating the midpoint between two towers as the full distance to each.

From the midpoint each tower is half the road away.

Mistake 04

Using the far angle as the near one.

Walking towards the tower raises the angle; the bigger angle is the closer point.

Mistake 05

Forgetting both triangles share the same tower height.

Write $h$ once and use it in both cotangent expressions.

Quick revision

Read this the night before the exam.

Same side: $h = \dfrac{d}{\cot\alpha - \cot\beta}$ .
Opposite sides: $h = \dfrac{d}{\cot\alpha + \cot\beta}$ .
$30^\circ$ and $60^\circ$ pair: difference $= \dfrac{2}{\sqrt{3}}$ , sum $= \dfrac{4}{\sqrt{3}}$ .
Two boats from a lighthouse: gap $= h(\cot\alpha - \cot\beta)$ .
Two towers from a midpoint: each is half the road away; heights $= \dfrac{d}{2}\tan\alpha$ and $\dfrac{d}{2}\tan\beta$ .
Foot and roof of a building: $d = \dfrac{b}{\tan\beta - \tan\alpha}$ , $H = d\tan\beta$ .

Practice: 15 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

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

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