Heights and Distances

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

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Compound figures: buildings, pedestals, broken objects

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⏱ 4 min read🧩 5 question types🎯 12 practice Q

The idea in one minute

Compound figures stack two heights or bend one object: a statue on a pedestal, a tower on a building, a broken tree.

The method never changes. Draw the two triangles separately. The lower triangle gives the ground distance. The upper triangle gives the extra height.

Most answers reduce to one line: extra height $= d(\tan\beta - \tan\alpha)$ .

Two heights stacked

A statue stands on a pedestal. From one point you see the pedestal's top at $\alpha$ and the statue's head at $\beta$ .

The lower triangle fixes the ground: $d = p\cot\alpha$ , where $p$ is the pedestal's height. The statue's height is then:

s = d(\tan\beta - \tan\alpha)

Pedestal 30 m, angles $45^\circ$ and $60^\circ$ : $d = 30$ , and $s = 30(\sqrt{3} - 1) = 30\sqrt{3} - 30$ m.

Rule: The difference of tangents times the ground distance gives the stacked part. Never multiply the whole height twice.

A tower standing on a building

Same picture, other way round. You know the building, or you know the tower; the two angles come from the same point.

Building 20 m seen at $30^\circ$ , tower top at $60^\circ$ , both from one point:

Ground: $d = 20\cot 30^\circ = 20\sqrt{3}$ m.
Tower: $d(\tan 60^\circ - \tan 30^\circ) = 20\sqrt{3} \times \dfrac{2}{\sqrt{3}} = 40$ m.

Tip: The building's triangle is $20 = d\tan 30^\circ$ ; the tower's is $20 + t = d\tan 60^\circ$ . Two equations, two unknowns.

Watching from a roof

From a building's roof you look down at a tower's foot and up at its top. Two angles, one roof.

Depression $\alpha$ of the foot: $d = b\cot\alpha$ where $b$ is the building's height.
Elevation $\beta$ of the top: tower $= b + d\tan\beta$ .

Building 10 m, depression $30^\circ$ , elevation $60^\circ$ : $d = 10\sqrt{3}$ m and tower $= 10 + 10\sqrt{3}\times\sqrt{3} = 40$ m.

Watch: The building's height is inside the second triangle too. Forgetting the extra $b$ is the most common slip here.

The broken tree

A tree breaks and the top touches the ground some distance away, making angle $\theta$ with the ground.

Let $x$ be the distance from the foot to the touching point:

Standing stump $= x\tan\theta$ .
Broken piece (the slanting hypotenuse) $= x\sec\theta = \dfrac{x}{\cos\theta}$ .
Original height $=$ stump $+$ broken piece.

At $x = 15$ m and $\theta = 30^\circ$ : stump $= 5\sqrt{3}$ m, broken piece $= \dfrac{15}{\sqrt{3}/2} = 10\sqrt{3}$ m, total $= 15\sqrt{3}$ m.

Example: A quick check: the broken piece must be longer than the stump, because it is the slanting side.

A slipping ladder

A ladder of fixed length $L$ rests at angle $\theta_1$ , then slips to $\theta_2$ . Both positions share $L$ :

Foot slides by $L\cos\theta_2 - L\cos\theta_1$ (the new angle is smaller, so its cosine is bigger).
Top drops by $L\sin\theta_1 - L\sin\theta_2$ .

Ladder 10 m from $60^\circ$ to $30^\circ$ : the foot moves $10\times\dfrac{\sqrt{3}}{2} - 10\times\dfrac{1}{2} = 5\sqrt{3} - 5 = 5(\sqrt{3}-1)$ m.

Tip: Length stays constant. That single fact links the two triangles.

Question types you will see

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

Type 1very common2 practice Q

Statue on a pedestal

How to spot it:

A statue or flagstaff on a pillar; angles to the pillar's top and to the statue's top.

s = d(\tan\beta - \tan\alpha),\qquad d = p\cot\alpha

Method

Use the lower angle with the pedestal height to find $d$ .
Compute $\tan\beta - \tan\alpha$ .
Multiply by $d$ for the statue's height.

Why it works:

The two angles share one ground distance, so the height difference is one tangent subtraction.

Try this

A statue stands on a pedestal 30 m tall. From a point on the ground, the angles of elevation of the top of the pedestal and the top of the statue are $45^\circ$ and $60^\circ$ . The height of the statue is:

Show solution

$d = 30\cot 45^\circ = 30$ m.
$\tan 60^\circ - \tan 45^\circ = \sqrt{3} - 1$ .
Statue $= 30(\sqrt{3}-1)$ m.

Answer

$30(\sqrt{3}-1)$ m

Type 2common2 practice Q

Tower standing on a building

How to spot it:

The angles to the building's top and the tower's top come from the same point.

b = d\tan\alpha,\qquad b + t = d\tan\beta

Method

Write the building's triangle and find $d$ .
Write the full-height triangle.
Subtract for the tower alone.

Why it works:

Two triangles share the ground distance, so the tower is the tangent difference times $d$ .

Try this

From a point on the ground, the top of a building 20 m high and the top of a tower on it are seen at $30^\circ$ and $60^\circ$ . The height of the tower is:

Show solution

$d = 20\cot 30^\circ = 20\sqrt{3}$ m.
Total height $= 20\sqrt{3}\tan 60^\circ = 60$ m.
Tower $= 60 - 20 = 40$ m.

Answer

40 m

Type 3common2 practice Q

Depression and elevation from a roof

How to spot it:

From a roof: looking down at a tower's foot and up at its top.

d = b\cot\alpha,\qquad H = b + d\tan\beta

Method

Turn the depression $\alpha$ into an elevation at the tower's foot.
Solve that triangle for $d$ .
Add the building's height to $d\tan\beta$ for the tower.

Why it works:

The roof height feeds both triangles: once as the opposite side, once as the base offset.

Try this

From the roof of a 10 m building, the angle of depression of the foot of a tower is $30^\circ$ and the angle of elevation of its top is $60^\circ$ . The height of the tower is:

Show solution

$d = 10\cot 30^\circ = 10\sqrt{3}$ m.
Height above roof $= 10\sqrt{3}\tan 60^\circ = 30$ m.
Tower $= 10 + 30 = 40$ m.

Answer

40 m

Type 4common3 practice Q

The broken tree

How to spot it:

A tree or pole breaks and the top touches the ground at a distance and angle.

\text{total} = x\tan\theta + \frac{x}{\cos\theta}

Method

Mark the distance $x$ from foot to touching point.
Stump $= x\tan\theta$ ; broken piece $= x\sec\theta$ .
Add them for the original height.

Why it works:

The standing part and the slanting broken part meet at the break height.

Try this

A tree breaks and its top touches the ground 15 m from the foot, making $30^\circ$ with the ground. The original height of the tree was:

Show solution

Stump $= 15\tan 30^\circ = 5\sqrt{3}$ m.
Broken piece $= \dfrac{15}{\cos 30^\circ} = \dfrac{15 \times 2}{\sqrt{3}} = 10\sqrt{3}$ m.
Total $= 15\sqrt{3}$ m.

Answer

$15\sqrt{3}$ m

Type 5occasional

A slipping ladder

How to spot it:

A ladder's angle falls; how far the foot slides or the top drops is asked.

L\cos\theta_2 - L\cos\theta_1

Method

Write both triangles with the same ladder length $L$ .
Foot distances: $L\cos\theta_1$ then $L\cos\theta_2$ .
Subtract for the slide (or use sines for the drop).

Why it works:

The length is constant, so the change in the ground angle converts straight into ground distance.

Try this

A 10 m ladder leaning against a wall at $60^\circ$ with the ground starts slipping until it makes $30^\circ$ . The foot of the ladder slides through a distance of:

Show solution

First foot distance $= 10\cos 60^\circ = 5$ m.
New foot distance $= 10\cos 30^\circ = 5\sqrt{3}$ m.
Slide $= 5\sqrt{3} - 5 = 5(\sqrt{3}-1)$ m.

Answer

$5(\sqrt{3}-1)$ m

Formula sheet

Stacked object (statue on pedestal)

s = d(\tan\beta - \tan\alpha)

d comes from the lower triangle: d = pedestal height x cot alpha.

Tower on a building

t = d(\tan\beta - \tan\alpha),\quad d = b\cot\alpha

From a roof: depression and elevation

d = b\cot\alpha,\qquad H = b + d\tan\beta

Broken tree

\text{stump} = x\tan\theta,\quad \text{broken} = \frac{x}{\cos\theta}

x = distance from the foot to where the top touches.

Shortcuts that save time

⚡ Subtract the tans, multiply once

For any stacked object, extra height $= d(\tan\beta - \tan\alpha)$ . Compute the tangent difference first, then one multiplication.

Example

A flagstaff on a pillar is seen from 20 m away. The pillar's top and the flagstaff's top are at $45^\circ$ and $60^\circ$ . Find the flagstaff's height.

Show solution

$\tan 60^\circ - \tan 45^\circ = \sqrt{3} - 1$ .
Flagstaff $= 20(\sqrt{3}-1)$ m.

Answer

$20(\sqrt{3}-1)$ m

⚡ Broken tree: tan plus sec

Total height $= x(\tan\theta + \sec\theta)$ where $x$ is the ground distance to the touching point. At $30^\circ$ that is $x\left(\dfrac{1}{\sqrt{3}} + \dfrac{2}{\sqrt{3}}\right) = x\sqrt{3}$ .

Example

A bamboo breaks and its top touches the ground 9 m from the foot at $30^\circ$ with the ground. The original height was:

Show solution

Total $= 9\left(\tan 30^\circ + \sec 30^\circ\right) = 9 \times \sqrt{3}$ .
$= 9\sqrt{3}$ m.

Answer

$9\sqrt{3}$ m

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Computing the statue's height as $d\tan\beta$ (the whole height) and stopping.

Subtract the lower tangent: statue $= d(\tan\beta-\tan\alpha)$ .

Mistake 02

Forgetting the building's height when the tower stands on it.

Tower-on-building: total $= b + t$ , so the tower alone needs the tangent difference.

Mistake 03

Treating the broken piece of a tree as vertical.

The broken piece is the slanting side: $x\sec\theta$ .

Mistake 04

Adding the two angles' tangents instead of subtracting.

The lower triangle's height must come off; subtract.

Mistake 05

Using the ladder's old angle after it slips.

Each ladder position is its own triangle; only the length $L$ is shared.

Quick revision

Read this the night before the exam.

Stacked object: extra height $= d(\tan\beta - \tan\alpha)$ .
Ground first: $d$ = lower height $\\times$ cot of the lower angle.
Tower on building: two equations, $b = d\\tan\\alpha$ and $b+t = d\\tan\\beta$ .
From a roof: $d = b\\cot\\alpha$ (depression), tower $= b + d\\tan\\beta$ .
Broken tree: stump $= x\\tan\\theta$ , broken $= x\\sec\\theta$ , total is their sum.
Slipping ladder: foot slides $L(\\cos\\theta_2 - \\cos\\theta_1)$ .

Practice: 12 questions

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

Topic test · 12 questions

Suggested time 11 min · wrong answers go to your mistake notebook automatically.

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