Heights and Distances

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

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Standard angles: 30°, 45°, 60°

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

The idea in one minute

Almost every exam question uses only $30^\circ$ , $45^\circ$ and $60^\circ$ . Learn their tangent values once and most questions become one multiplication.

The cotangent column is the handiest: it turns a height into a distance. At $45^\circ$ the distance equals the height; at $30^\circ$ it is $\sqrt{3}$ times the height; at $60^\circ$ it is $\sqrt{3}$ divided by the height's factor.

The three friendly angles

Exam papers love $30^\circ$ , $45^\circ$ and $60^\circ$ because their ratios are clean.

Angle	$\tan$	$\cot$ (distance per unit height)
$30^\circ$	$\dfrac{1}{\sqrt{3}}$	$\sqrt{3}$
$45^\circ$	$1$	$1$
$60^\circ$	$\sqrt{3}$	$\dfrac{1}{\sqrt{3}}$

Tip: Memorise the $\cot$ column as "how far you stand from a 1 m object". At $30^\circ$ you stand $\sqrt{3}$ m; at $60^\circ$ only $\dfrac{1}{\sqrt{3}}$ m.

Worked number: a tower is $30\sqrt{3}$ m tall and its top is seen at $30^\circ$ . Distance $= 30\sqrt{3} \times \sqrt{3} = 90$ m.

Swapping between height and distance

One multiplication is enough, once the angle is standard.

d = h \times \cot\theta \qquad h = d \times \tan\theta

A point is 90 m from the foot of a tower, and the top is at $60^\circ$ . Height $= 90 \times \sqrt{3} = 90\sqrt{3}$ m. Check the size: a near angle ( $60^\circ$ ) with a long distance must give a tall tower.

Rule: Bigger angle, closer point. At $60^\circ$ the tower is about 1.7 times the distance; at $30^\circ$ it is about 0.6 times.

Spotting the angle from the sides

Sometimes the sides are given and the angle is asked.

A tower 20 m tall casts a shadow 20 m long. Height equals distance, and only one angle makes the legs equal: $45^\circ$ .

Example: A tower is $\sqrt{3}$ times as far from you as it is tall, so $\cot\theta = \sqrt{3}$ , so $\theta = 30^\circ$ .

Ladders and slanting wires

When the slanting length $L$ is given, use sine and cosine.

A 20 m ladder leans on a wall at $30^\circ$ with the ground:

Height on the wall $= 20 \times \dfrac{1}{2} = 10$ m.
Foot from the wall $= 20 \times \dfrac{\sqrt{3}}{2} = 10\sqrt{3}$ m.

Watch: The $30^\circ$ angle is between the ladder and the ground. If it is given at the wall, swap sine and cosine.

The fifteen and seventy-five pair

Two more exact values appear in harder papers:

\tan 75^\circ = 2 + \sqrt{3}, \qquad \tan 15^\circ = 2 - \sqrt{3}

They multiply to 1, because $15^\circ + 75^\circ = 90^\circ$ .

A tower 10 m tall is seen at $15^\circ$ . Distance $= \dfrac{10}{2-\sqrt{3}} = 10(2+\sqrt{3}) = 20 + 10\sqrt{3}$ m. Rationalise by multiplying top and bottom by $2+\sqrt{3}$ .

When the sun moves

As the sun climbs, the shadow shrinks. Two positions of the same shadow give one equation.

A tower's shadow shortens by 10 m as the sun moves from $30^\circ$ to $60^\circ$ :

h(\cot 30^\circ - \cot 60^\circ) = 10 \Rightarrow h \times \frac{2}{\sqrt{3}} = 10 \Rightarrow h = 5\sqrt{3} \text{ m}

Tip: For the $30^\circ$ and $60^\circ$ pair, $\cot 30^\circ - \cot 60^\circ = \dfrac{2}{\sqrt{3}}$ . The same difference appears in many questions.

Question types you will see

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

Type 1very common4 practice Q

Height or distance at a standard angle

How to spot it:

One standard angle with one side; the other side is asked.

d = h\cot\theta

Method

Pick the standard angle's cot value.
Multiply the height by it for the distance, or divide for the height.
Keep $\sqrt{3}$ exact.

Why it works:

A standard angle fixes the ratio of the two legs, so one multiplication finishes it.

Try this

The angle of elevation of the top of a tower $30\sqrt{3}$ m high, from a point on the ground, is $30^\circ$ . The distance of the point from the foot of the tower is:

Show solution

$d = 30\sqrt{3} \times \cot 30^\circ$ .
$d = 30\sqrt{3} \times \sqrt{3}$ .
$d = 90$ m.

Answer

90 m

Type 2common2 practice Q

Find the angle from the sides

How to spot it:

Both legs are given (or their ratio) and the sun's or tower's angle is asked.

\tan\theta = \frac{h}{d}

Method

Form the ratio height over distance.
Reduce it to a surd form.
Match against the standard tangent table.

Why it works:

The legs decide the angle, so the ratio reads the angle straight off the table.

Try this

A vertical tower 20 m tall casts a shadow 20 m long on the ground. The sun's elevation is:

Show solution

$\tan\theta = \dfrac{20}{20} = 1$ .
$\tan 45^\circ = 1$ .
$\theta = 45^\circ$ .

Answer

$45^\circ$

Type 3common2 practice Q

Ladder against a wall

How to spot it:

A ladder's length and its angle with the ground are given.

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

Method

Confirm the angle is with the ground, not the wall.
Height reached $= L\sin\theta$ .
Foot distance $= L\cos\theta$ .

Why it works:

The ladder is the hypotenuse, so sine gives the wall side and cosine the ground side.

Try this

A 20 m ladder leans against a vertical wall, making $30^\circ$ with the ground. The height reached on the wall and the distance of the foot from the wall are:

Show solution

Height $= 20\sin 30^\circ = 20 \times \dfrac{1}{2} = 10$ m.
Distance $= 20\cos 30^\circ = 20 \times \dfrac{\sqrt{3}}{2}$ .
Distance $= 10\sqrt{3}$ m.

Answer

10 m and $10\sqrt{3}$ m

Type 4occasional2 practice Q

Fifteen or seventy-five degrees

How to spot it:

An uncommon angle that is 15 or 75 degrees; options contain surds.

\tan 75^\circ = 2+\sqrt{3},\qquad \tan 15^\circ = 2-\sqrt{3}

Method

Replace the tangent with its surd value.
Rationalise the denominator with the partner surd.
The two surds multiply to 1.

Why it works:

These two angles have exact surd tangents, so the arithmetic stays exact.

Try this

From a point on the ground, the angle of elevation of the top of a 10 m tower is $15^\circ$ . The distance of the point from the foot of the tower is:

Show solution

$d = 10 \times \cot 15^\circ = \dfrac{10}{2-\sqrt{3}}$ .
Multiply top and bottom by $2+\sqrt{3}$ : $d = 10(2+\sqrt{3})$ .
$d = 20 + 10\sqrt{3}$ m.

Answer

$20+10\sqrt{3}$ m

Type 5occasional

Shadow changing as the sun moves

How to spot it:

The same object's shadow is described at two sun positions; the change in shadow is given.

h(\cot\alpha - \cot\beta) = \text{change in shadow}

Method

Write both shadow lengths: $h\cot\alpha$ and $h\cot\beta$ .
Subtract and equate to the given change.
Solve for $h$ .

Why it works:

Each sun position makes its own triangle with the same height, so one equation links them.

Try this

As the sun rises from $30^\circ$ to $60^\circ$ , the shadow of a tower shortens by 10 m. The height of the tower is:

Show solution

Shadow at $30^\circ$ : $h\sqrt{3}$ . Shadow at $60^\circ$ : $\dfrac{h}{\sqrt{3}}$ .
$h\sqrt{3} - \dfrac{h}{\sqrt{3}} = \dfrac{2h}{\sqrt{3}} = 10$ .
$h = 5\sqrt{3}$ m.

Answer

$5\sqrt{3}$ m

Formula sheet

Tangent values

\tan 30^\circ = \frac{1}{\sqrt{3}},\quad \tan 45^\circ = 1,\quad \tan 60^\circ = \sqrt{3}

Distance from height

d = h\cot\theta

cot 30 = sqrt3, cot 45 = 1, cot 60 = 1/sqrt3.

Ladder on a wall

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

Theta is the ladder's angle with the ground.

Fifteen and seventy-five

\tan 15^\circ = 2-\sqrt{3},\qquad \tan 75^\circ = 2+\sqrt{3}

Shortcuts that save time

⚡ Read distance straight off the cot column

With a standard angle, distance = height $\times$ the cot value. No division, no fraction juggling.

Example

A tower 30 m tall has its top seen at $60^\circ$ from a point on the ground. How far is the point from the foot?

Show solution

$\cot 60^\circ = \dfrac{1}{\sqrt{3}}$ .
Distance $= 30 \times \dfrac{1}{\sqrt{3}} = \dfrac{30}{\sqrt{3}} = 10\sqrt{3}$ m.

Answer

$10\sqrt{3}$ m

⚡ Fifteen and seventy-five multiply to one

Angles that add to $90^\circ$ have tangents that multiply to 1. Use this to check answers or flip a division into a multiplication.

Example

A tower is seen at $75^\circ$ from 10 m away. Using $\tan 15^\circ = 2-\sqrt{3}$ , its height is:

Show solution

$h = 10 \times \tan 75^\circ = 10 \times \dfrac{1}{2-\sqrt{3}}$ .
$= 10(2+\sqrt{3}) = 20 + 10\sqrt{3}$ m.

Answer

$20+10\sqrt{3}$ m

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Swapping the values: writing $\tan 30^\circ = \sqrt{3}$ .

$\tan 30^\circ = \dfrac{1}{\sqrt{3}}$ ; the big value belongs to $60^\circ$ .

Mistake 02

Computing distance as height $\times\tan\theta$ .

Distance $= h \times \cot\theta$ ; the tangent gives height from distance.

Mistake 03

Using the ladder's length as the height it reaches.

Height on the wall $= L\sin\theta$ ; the ladder itself is the slanting side.

Mistake 04

Placing the $30^\circ$ ladder angle at the top of the wall.

The ladder's angle is normally with the ground. At the wall, the angle is $60^\circ$ .

Mistake 05

Replacing $\sqrt{3}$ by 1.7 in an exact-answer question.

Keep $\sqrt{3}$ symbolic unless the question says approximately or gives $1.73$ .

Quick revision

Read this the night before the exam.

$\tan 30^\circ = \dfrac{1}{\sqrt{3}}$ , $\tan 45^\circ = 1$ , $\tan 60^\circ = \sqrt{3}$ .
$\cot$ column: $\sqrt{3}$ , $1$ , $\dfrac{1}{\sqrt{3}}$ for $30^\circ$ , $45^\circ$ , $60^\circ$ .
$d = h\cot\theta$ and $h = d\tan\theta$ .
Ladder: height $= L\sin\theta$ , foot distance $= L\cos\theta$ .
$\tan 75^\circ = 2+\sqrt{3}$ , $\tan 15^\circ = 2-\sqrt{3}$ , product $1$ .
Shadow change: $h(\cot\alpha - \cot\beta)$ = change in length.

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

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