Heights and Distances

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

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Moving observers: speed and time

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⏱ 4 min read🧩 5 question types🎯 13 practice Q

The idea in one minute

A car, a man or a plane moves while someone watches. Motion adds only one new fact: distance = speed $\times$ time.

That product replaces the 'walked distance' of the two-angle questions. So the master equation becomes speed $\times$ time $= h(\cot\alpha - \cot\beta)$ .

Convert every speed to metres per second first. Then the question is an old friend.

Motion only supplies the ground distance

In a moving-observer question, the triangle does not change. Only the ground between the two viewpoints is now covered at a speed.

v \times t = h(\cot\alpha - \cot\beta)

Here $v$ is the speed, $t$ the time between the two angle readings, $h$ the tower height, $\alpha$ the first (farther) angle and $\beta$ the second (nearer) angle.

Example: A car running at 6 m/s takes 6 s to change the elevation of a tower top from $30^\circ$ to $60^\circ$ . Ground covered $= 36$ m. Then $36 = h \times \dfrac{2}{\sqrt{3}}$ , so $h = 18\sqrt{3}$ m.

Finding the height

Given speed, time and both angles, the height comes out in two lines.

Ground distance $= v \times t$ .
Divide by $\cot\alpha - \cot\beta$ .

At $4$ m/s for $5$ s the car covers 20 m, so $h = 20 \times \dfrac{\sqrt{3}}{2} = 10\sqrt{3}$ m.

Tip: For $30^\circ \to 60^\circ$ , $h = (v t)\times\dfrac{\sqrt{3}}{2}$ . One multiplication, no surd subtraction.

Finding the speed or the time

The same equation reads in any direction. Solve for what is missing.

A car changes the elevation from $30^\circ$ to $60^\circ$ in 5 s, and the tower is known to be $10\sqrt{3}$ m. Ground covered $= h \times \dfrac{2}{\sqrt{3}} = 20$ m. Speed $= 20 \div 5 = 4$ m/s.

Watch: Speeds in km/h must be converted: multiply by $\dfrac{5}{18}$ to get m/s. 36 km/h is 10 m/s.

Time to reach the foot

Sometimes the question asks how much longer the car needs to reach the tower.

After the second reading, the car is still $h\cot\beta$ away (using the nearer angle). Divide that distance by the speed.

A tower is $18\sqrt{3}$ m tall and a car sees its top at $30^\circ$ . Distance $= 18\sqrt{3} \times \sqrt{3} = 54$ m. At 6 m/s the car needs $54 \div 6 = 9$ s more.

Moving on both sides

Two friends drive towards the same tower from opposite sides at the same speed.

Each starts at $h\cot\alpha$ or $h\cot\beta$ away. Their arrival times differ by:

\Delta t = \frac{h(\cot\alpha - \cot\beta)}{v}

Tower $30\sqrt{3}$ m, angles $60^\circ$ and $30^\circ$ , both at 6 m/s: distances $30$ m and $90$ m, times $5$ s and $15$ s. The nearer starter wins by 10 s.

Rising straight up

A balloon rises vertically while someone watches a point on the ground. The horizontal distance stays fixed; the height changes.

At horizontal distance $d$ : heights are $d\tan 30^\circ = \dfrac{d}{\sqrt{3}}$ and $d\tan 60^\circ = d\sqrt{3}$ . The rise $= d \times \dfrac{2}{\sqrt{3}}$ . With $d = 30$ m, the balloon climbed $20\sqrt{3}$ m.

Rule: Horizontal motion changes the distance; vertical motion changes the height. Check which one is fixed before writing tan or cot.

Question types you will see

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

Type 1very common5 practice Q

Find the height from speed and time

How to spot it:

A vehicle moves at a given speed for a given time while the elevation angle changes.

h = \frac{v\,t}{\cot\alpha - \cot\beta}

Method

Compute the ground covered: $v \times t$ .
Write $\cot\alpha - \cot\beta$ for the two angles.
Divide to get $h$ .

Why it works:

The covered ground equals the difference of the two cotangent distances of the same height.

Try this

A car moving at 6 m/s takes 6 seconds to change the angle of elevation of a tower's top from $30^\circ$ to $60^\circ$ . The height of the tower is:

Show solution

Ground $= 6 \times 6 = 36$ m.
$\cot 30^\circ - \cot 60^\circ = \dfrac{2}{\sqrt{3}}$ .
$h = 36 \times \dfrac{\sqrt{3}}{2} = 18\sqrt{3}$ m.

Answer

$18\sqrt{3}$ m

Type 2common3 practice Q

Find the speed or the time

How to spot it:

The height and both angles are known; a speed or a time is asked.

v = \frac{h(\cot\alpha - \cot\beta)}{t}

Method

Compute the ground: $h(\cot\alpha - \cot\beta)$ .
Divide by the time for the speed, or by the speed for the time.
Convert units if the options are in km/h.

Why it works:

With the height known, the chain gives the ground and speed is ground over time.

Try this

The angle of elevation of the top of a $10\sqrt{3}$ m tower changes from $30^\circ$ to $60^\circ$ as a car approaches in 5 seconds. The speed of the car is:

Show solution

Ground $= 10\sqrt{3} \times \dfrac{2}{\sqrt{3}} = 20$ m.
Speed $= 20 \div 5 = 4$ m/s.
In km/h: $4 \times \dfrac{18}{5} = 14.4$ km/h.

Answer

4 m/s (14.4 km/h)

Type 3common2 practice Q

Time to reach the foot of the tower

How to spot it:

After the angle change the question asks how much longer the mover needs to reach the tower.

t = \frac{h\cot\beta}{v}

Method

Use the angle at the mover's current position.
Remaining distance $= h\cot\beta$ .
Divide by the speed.

Why it works:

One triangle is enough: the current angle fixes the remaining ground.

Try this

A tower is $18\sqrt{3}$ m tall. A car approaching it at 6 m/s currently sees the top at $30^\circ$ . The time the car needs to reach the foot of the tower is:

Show solution

Distance $= 18\sqrt{3} \times \cot 30^\circ = 18\sqrt{3} \times \sqrt{3} = 54$ m.
Time $= 54 \div 6$ .
$= 9$ seconds.

Answer

9 seconds

Type 4occasional2 practice Q

Two movers on opposite sides

How to spot it:

Two people drive towards the same tower from opposite sides; who arrives first and by how much.

\Delta t = \frac{h(\cot\alpha - \cot\beta)}{v}

Method

Find each starting distance: $h\cot\alpha$ and $h\cot\beta$ .
Divide each by its speed to get two times.
Subtract for the gap.

Why it works:

Each mover runs an independent race against their own cotangent distance.

Try this

A tower $30\sqrt{3}$ m tall stands between two friends who see its top at $60^\circ$ and $30^\circ$ . Both drive towards the tower at 6 m/s, starting together. By how many seconds does the first one reach the tower before the second?

Show solution

Friend with $60^\circ$ : distance $= 30\sqrt{3} \times \dfrac{1}{\sqrt{3}} = 30$ m, time $= 5$ s.
Friend with $30^\circ$ : distance $= 30\sqrt{3} \times \sqrt{3} = 90$ m, time $= 15$ s.
Gap $= 15 - 5 = 10$ s.

Answer

10 seconds

Type 5occasional

Observer rising vertically

How to spot it:

A balloon or lift rises; the angle of depression of a fixed point changes.

\text{rise} = d(\tan\beta - \tan\alpha)

Method

Fix the horizontal distance $d$ to the point.
Two heights: $d\tan\alpha$ (start) and $d\tan\beta$ (later).
Subtract for the rise.

Why it works:

Only the height changes, so the rise is the difference of two tangents over the same base.

Try this

A balloon rises straight up. The angle of depression of a stone 30 m away (horizontally) changes from $30^\circ$ to $60^\circ$ . The height the balloon gained is:

Show solution

Start: $30\tan 30^\circ = \dfrac{30}{\sqrt{3}} = 10\sqrt{3}$ m.
Later: $30\tan 60^\circ = 30\sqrt{3}$ m.
Rise $= 30\sqrt{3} - 10\sqrt{3} = 20\sqrt{3}$ m.

Answer

$20\sqrt{3}$ m

Formula sheet

Moving observer chain

v\,t = h(\cot\alpha - \cot\beta)

alpha = first (farther) angle, beta = second (nearer) angle.

Time to reach the foot

t = \frac{h\cot\beta}{v}

Use the angle at the car's current position.

Vertical rise

\text{rise} = d(\tan\beta - \tan\alpha)

d = fixed horizontal distance; angles of depression shrink as the balloon rises.

Speed conversion

1\ \text{km/h} = \frac{5}{18}\ \text{m/s}

Shortcuts that save time

⚡ One chain, any unknown

$v t = h(\cot\alpha - \cot\beta)$ contains every moving-observer question. Cover the unknown and solve.

Example

A jeep takes 3 s at 8 m/s to change a tower's elevation from $30^\circ$ to $60^\circ$ . Find the height.

Show solution

Ground $= 8 \times 3 = 24$ m.
$h = 24 \times \dfrac{\sqrt{3}}{2} = 12\sqrt{3}$ m.

Answer

$12\sqrt{3}$ m

⚡ Check the units before the triangle

Convert km/h to m/s with $5/18$ before anything else. A speed in the wrong unit spoils an otherwise perfect triangle.

Example

A car at 36 km/h takes 10 s to change the elevation from $30^\circ$ to $60^\circ$ . Find the height.

Show solution

$36$ km/h $= 36 \times \dfrac{5}{18} = 10$ m/s.
Ground $= 10 \times 10 = 100$ m.
$h = 100 \times \dfrac{\sqrt{3}}{2} = 50\sqrt{3}$ m.

Answer

$50\sqrt{3}$ m

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using $v\times t$ as the height instead of the ground distance.

Speed times time is ground covered; it equals the cotangent difference.

Mistake 02

Working in km/h with metres in the same equation.

Convert with $\times\dfrac{5}{18}$ first.

Mistake 03

Giving the total time when the remaining time to the foot is asked.

After the second angle, the car still has $h\cot\beta$ to cover; divide that by the speed.

Mistake 04

Adding angles of two observers who move on opposite sides.

Opposite sides still add the two cotangent distances; only the ground differs.

Mistake 05

Using cot when the observer rises vertically.

With a fixed horizontal distance, both readings use tan with the same $d$ .

Quick revision

Read this the night before the exam.

$v\,t = h(\cot\alpha - \cot\beta)$ : the master chain.
$30^\circ \to 60^\circ$ : $h = (v t)\times\dfrac{\sqrt{3}}{2}$ .
Remaining time to the foot: $h\cot\beta \div v$ .
km/h to m/s: multiply by $\dfrac{5}{18}$ .
Opposite-side movers: distances $h\cot\alpha$ and $h\cot\beta$ , times differ by their difference over $v$ .
Balloon rising: rise $= d(\tan\beta - \tan\alpha)$ with $d$ fixed.

Practice: 13 questions

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

Topic test · 13 questions

Suggested time 12 min · wrong answers go to your mistake notebook automatically.

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