Heights and Distances

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

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Subtopic 4 of 5·← Two observation points (two angles)·Compound figures: buildings, pedestals, broken objects →

Moving observers: speed and time

When a car/boat/man moves between the two observation points, the distance walked is speed × time. Chain:

$\text{speed}\times\text{time}=h(\cot\theta_1-\cot\theta_2)$

then solve for $h$ (or for the speed if $h$ is given). Keep angles straight: moving toward the object the angle increases; moving away it decreases.

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 13 practice Q

Detailed notes

Speed and time become distance

When a car, boat or man moves between the two observation points, the walked distance is just speed × time. Chain it to the cotangent formula: $v \times t = h(\cot\theta_1 - \cot\theta_2)$ and every "moving observer" question is a two-angle question in a coat. Keep units married: km/h needs hours, m/s needs seconds; $36$ km/h $= 10$ m/s is the conversion to have ready.

The four variants

Find h: distance walked given via $v, t$ ; angles given → $h = \frac{vt}{\cot\theta_1 - \cot\theta_2}$ .
Find v or t: h given; compute the cot-gap and divide/multiply.
Depression from above: a car approaches a tower/lighthouse; depressions shrink as it nears? No — as the car approaches, the depression grows (it looks steeper down). Moving away shrinks it. The larger angle is always the nearer position.
Time to reach the base: after the angle reaches $\beta$ , the remaining distance is $h\cot\beta$ ; the time to the base is $\frac{h\cot\beta}{v}$ .

Direction bookkeeping

Approaching: angle increases ( $\theta_2 > \theta_1$ ), distance walked $= h(\cot\theta_1 - \cot\theta_2)$ .
Receding: angle decreases, walked $= h(\cot\theta_2 - \cot\theta_1)$ — same formula, angles swapped. The identity "nearer ⇒ steeper" kills half the sign errors: if your formula gives a negative walked distance, you assigned the angles backwards.

Boats and streamers

A boat moving toward a cliff at $u$ m/s sees the depression grow from $\alpha$ to $\beta$ in $t$ seconds: the boat covered $ut$ metres, so $ut = h(\cot\alpha - \cot\beta)$ . The distance from the cliff after the sighting is $h\cot\beta$ ; the total time to reach the cliff base is $\frac{h\cot\beta}{u}$ — a favourite "how much longer" add-on.

Worked micro-example (every variant in one)

A $30\sqrt3$ m tower: a car's depression changes from $30^\circ$ to $60^\circ$ in $10$ s at constant speed. Walked: $30\sqrt3(\cot30 - \cot60) = 30\sqrt3 \cdot \frac{2}{\sqrt3} = 60$ m → speed $6$ m/s. Remaining to the base: $30\sqrt3\cot60 = 30$ m → $5$ s more. Total from first sighting: $15$ s. Reverse variants: given speed 6, find h; given h and the times, find the angles' gap; given everything, the time to the base. Same skeleton — four different questions.

Quick revision

$vt = h(\cot\theta_{\text{far}} - \cot\theta_{\text{near}})$ : far angle is the SMALLER one.
30–60 pair: $vt = \frac{2h}{\sqrt3}$ , so $h = \frac{vt\sqrt3}{2}$ .
After the second sighting, the remaining distance is $h\cot\theta_{\text{near}}$ .
Convert units first: km/h → m/s is ×5/18.
Nearer ⇒ larger angle; negative answers mean swapped angles.

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: Car/boat approach: find the heightvery common5 practice Q

How to spot it:

'Moving at v m/s for t s the depression/elevation changes from α to β' — height of the tower/lighthouse asked.

$h = \frac{vt}{\cot\alpha - \cot\beta}$ with $\alpha$ the smaller (farther) angle; 30–60 pair → $h = \frac{vt\sqrt3}{2}$.

Compute the walked distance $vt$ (units consistent!).
Apply the same-side two-angle formula.
Sanity: h should be a little less than $vt\sqrt3$ for the 30–60 pair.

Why: the motion only manufactures the walked distance; the trigonometry is the standard pair.

Example: From the top of a tower the depression of a car changes from $30^\circ$ to $60^\circ$ as the car approaches at $6$ m/s for $6$ s. The tower's height is:

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

Type 2: Reverse: find the speed or timecommon3 practice Q

How to spot it:

Height given; the speed (or time) for the angle to change from α to β asked.

Walked distance $= h(\cot\alpha - \cot\beta)$; then $v = \frac{\text{distance}}{t}$ or $t = \frac{\text{distance}}{v}$.

Compute the cot-gap distance first.
Divide by the given time (for speed) or speed (for time).
"How much longer to reach the base" adds $\frac{h\cot\beta}{v}$ .

Why: the geometry fixes the distance; rate arithmetic finishes it.

Example: A car travelling at $4$ m/s takes $5$ s for the elevation of a tower top to rise from $30^\circ$ to $60^\circ$ . The tower's height is:

$vt = 20$ : $h \cdot \frac{2}{\sqrt3} = 20 \Rightarrow h = 10\sqrt3$ m.

Type 3: Time to reach the foot of the objectoccasional2 practice Q

How to spot it:

After the second angle is observed, how much longer until the observer is at the base — or the total journey time.

Remaining distance $= h\cot\beta$ (the near angle); time $= \frac{h\cot\beta}{v}$. Total time adds the elapsed $t$.

Find h from the first leg.
The remaining leg is the near distance $h\cot\beta$ .
Divide by v; add elapsed time only if "total" is asked.

Why: the near cotangent distance IS what is left to travel — no new trigonometry needed.

Example: A car at $6$ m/s watched from a $18\sqrt{3}$ m tower sees its depression grow from $30^\circ$ to $60^\circ$ . After the second sighting, the time to reach the tower base is:

Remaining $= 18\sqrt3\cot60^\circ = 18$ m $\Rightarrow \frac{18}{6} = 3$ s.

Type 4: Two moving observers / positions from motionoccasional2 practice Q

How to spot it:

Two cars/boats approach the same tower on one road, or one object moves while a fixed position is tracked.

Each observer owns a cot-distance from the base; the gap between two approaching vehicles closes at the SUM of speeds.

Write each position as a cot product of h where angles are known.
For two vehicles approaching each other, the gap closes at (u + v).
Solve the linear condition stated (meeting time, gap, who is where).

Why: the trigonometry pins positions; relative-speed arithmetic tracks their change — two separate layers.

Example: Two cars approach a tower from the same road. When the slower car is $60\sqrt{3}$ m away, its elevation of the tower top is $30^\circ$ . The tower's height is:

$h = 60\sqrt3\tan30^\\circ = 60$ m (speeds matter for later time questions, not for h).

Formulas

Distance walked

\text{distance}=\text{speed}\times\text{time}

Chain to height

v\,t=h(\cot\theta_1-\cot\theta_2)

Shortcut tricks

⚡ Convert speed and time first

Reduce to a plain distance (e.g. $6\times6=36$ m) and the problem becomes the standard two-angle pattern.

Example: From a tower top, a car's angle of depression changes from $30^\circ$ to $60^\circ$ as the car approaches at a uniform 6 m/s for 6 s. Find the tower's height.

$36=h(\cot30^\circ-\cot60^\circ)=h\cdot\frac2{\sqrt3}\Rightarrow h=18\sqrt3$ m.

⚡ Read it backwards when height is given

If $h$ is known and the speed is asked, compute the gap $h(\cot\theta_1-\cot\theta_2)$ first, then divide by time.

Example: A car approaching a tower at a steady 4 m/s takes 5 s for the angle of elevation of the tower top to change from $30^\circ$ to $60^\circ$ . Find the tower's height.

Gap $=4\times5=20=h\cdot\frac2{\sqrt3}\Rightarrow h=10\sqrt3$ m.

Where students lose marks

Mixing units (km/h with seconds) — convert before chaining.
Assigning the larger angle to the farther position.

Practice sets — 13 questions

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

Topic test · 13 questions

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