MathsICSE Class 10 · Heights and Distances

Two Vertical Objects: Building and Tower Problems in Heights and Distances (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

With two vertical objects, draw a horizontal line from the observer's eye across to the other object. It splits the figure into two right triangles: the depression of the foot gives the gap $d = H\cot\beta$, and the elevation of the top gives the extra height $d\tan\alpha$, added to (or subtracted from) $H$.

Draw the horizontal from your eye: the foot's depression β gives d, and then d tan α is the extra height added to H.

Illustration drawn for QuestionBuddy.

Understand Two Vertical Objects

Two buildings, a building and a tower, a tower with a flagstaff, or two poles across a road: each is a 2-D problem with two right triangles, the limit the CISCE 2027 syllabus allows. The skill is drawing the extra horizontal line that creates them. 2025 Q6(ii) was this type.

Type 1: from the top of a building, elevation of the tower top and depression of its foot. Let the building be $H$ high and the gap $d$ .

Depression $β$ of the foot: $tan β = \frac{H}{d}$ , so $d = H cot β$ .
Draw the horizontal from the building top to the tower. Elevation $α$ of the tower top: extra height $= d tan α$ .
Tower height $= H + d tan α$ .

Type 2: from the top of a tall building, depressions of the top and foot of a shorter tower. The depression of the foot gives $d$ ; the depression $α$ of the top gives the drop $d tan α$ . Tower $= H - d tan α$ .

Type 3: a flagstaff (or statue, mast, billboard) on a building, from a ground point. At distance $d$ , elevation $α$ of the building top and $β$ of the flagstaff top: building $= d tan α$ , top $= d tan β$ , flagstaff $= d (tan β - tan α)$ .

Type 4: two poles on either side of a road, seen from a point between them. If the poles have equal height $h$ and the road is $w$ wide: $h = x tan α = (w - x) tan β$ . Solve for $x$ , then $h$ .

Type 5: from the foot of each, the elevation of the other. These are two separate triangles sharing the base $d$ : heights $d tan α$ and $d tan β$ .

When the gap is given (as in 2025 Q6(ii)). If the horizontal distance is known, each angle gives a vertical length directly: $100 tan 3 5^{\circ}$ and $100 tan 1 4^{\circ}$ . Then combine the lengths along the building.

Two Vertical Objects: formulas to remember

Gap from depression of the foot: $d = H cot β$; $H$ = height of the observer's building.
Taller tower (elevation of its top): $tower = H + d tan α$; The horizontal from the eye splits the tower.
Shorter tower (depression of its top): $tower = H - d tan α$; Subtract the drop from $H$ .
Flagstaff on a building: $flagstaff = d (tan β - tan α)$; Both angles from the same ground point.
Equal poles across a road of width w: $x tan α = (w - x) tan β$; $x$ = distance from the first pole.

How to solve Two Vertical Objects questions

Find the triangle that has two known parts, solve it for the gap, then use the gap in the second triangle.

1Draw both objects vertically on one ground line, with the gap $d$ between their feet.
2From the observer's eye, draw a dashed horizontal across to the other object and label where it meets.
3Mark each angle from that horizontal: elevation above it, depression below it.
4Solve the triangle with two known parts first (usually the depression of a foot, which gives $d$ ).
5Use $d$ in the second triangle to get the vertical piece above or below the horizontal.
6Add or subtract to get the required height; round only the final answer.

Two Vertical Objects in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Height of a water tank tower from a school roof

Suppose you are on the flat roof of your school, 15 m above the ground, and see the top of a water tank tower at an angle of elevation of $2 5^{\circ}$ and its foot at an angle of depression of $4 0^{\circ}$ .

How far away is the tower, and how tall is it, to 2 decimal places? ( $tan 2 5^{\circ} = 0.4663$ , $tan 4 0^{\circ} = 0.8391$ )

Try it first, then show the solutionSolution

Gap $d = \frac{15}{t a n 4 0 ^{\circ}} = \frac{15}{0.8391} \approx 17.88$ m.
Height above the roof level $= 17.876 \times 0.4663 \approx 8.34$ m.
Tower $= 15 + 8.336 \approx 23.34$ m.

Answer: About 17.88 m away and 23.34 m tall

Memory hook: Down to the foot gives the gap; up to the top gives the extra.

Size of a rooftop billboard

Imagine standing on a road 50 m from a building. The roof edge is at an angle of elevation of $2 8^{\circ}$ and the top of a billboard standing on the roof at $3 5^{\circ}$ .

How tall is the billboard, to 2 decimal places? ( $tan 2 8^{\circ} = 0.5317$ , $tan 3 5^{\circ} = 0.7002$ )

Try it first, then show the solutionSolution

Roof height $= 50 \times 0.5317 = 26.585$ m; billboard top $= 50 \times 0.7002 = 35.01$ m.
Billboard $= 35.01 - 26.585 \approx 8.43$ m.

Answer: About 8.43 m

Memory hook: Same point, two angles: subtract the two heights.

Tricks to remember Two Vertical Objects

Short memory aids. Say each one aloud once, then use it on a question the same day.

One horizontal line does it
From the observer's eye, draw one dashed horizontal across to the other object. Everything above it is 'elevation', everything below is 'depression'.
Use it when: Turning any two-object figure into two right triangles.
Foot first
The depression of the foot uses the full building height, so it always gives the gap $d$ first.
Use it when: Choosing the starting triangle.
Taller adds, shorter subtracts
Looking up at the other top: add to $H$ . Looking down at it: subtract from $H$ .
Use it when: Getting the final height right.

Two Vertical Objects: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Equal angles, standard value · 1 mark · MCQBoard level

From the top of a building 10 m high, the angle of elevation of the top of a tower is $4 5^{\circ}$ and the angle of depression of its foot is $4 5^{\circ}$ . The height of the tower is:
(a) 10 m
(b) 15 m
(c) 20 m
(d) $102$ m

Try it first, then show the solutionSolution

Gap $= 10 cot 4 5^{\circ} = 10$ m; extra height $= 10 tan 4 5^{\circ} = 10$ m.
Tower $= 10 + 10 = 20$ m.

Answer: (c)

Example 2: Two equal poles across a road · 4 marks · Short answerBoard level

Two poles of equal height stand on opposite sides of a road 80 m wide. From a point on the road between them, the angles of elevation of their tops are $6 0^{\circ}$ and $3 0^{\circ}$ . Find the height of the poles, to 2 decimal places, and the distance of the point from each pole.

Try it first, then show the solutionSolution

Let the point be $x$ m from the first pole (angle $6 0^{\circ}$ ), so $80 - x$ m from the second.
$h = x tan 6 0^{\circ} = 3 x$ and $h = (80 - x) tan 3 0^{\circ} = \frac{80 - x}{3}$ .
$3 x = 80 - x$ , so $x = 20$ m and $80 - x = 60$ m.
$h = 203 \approx 34.64$ m.

Answer: Height ≈ 34.64 m; 20 m from one pole, 60 m from the other

Example 3: Elevation of the top, depression of the foot · 4 marks · Short answerHarder board level

From the top of a building 18 m high, the angle of elevation of the top of a tower is $3 4^{\circ}$ and the angle of depression of its foot is $2 6^{\circ}$ . Find the distance between the building and the tower, and the height of the tower, to 2 decimal places. ( $tan 3 4^{\circ} = 0.6745$ , $tan 2 6^{\circ} = 0.4877$ )

Try it first, then show the solutionSolution

Depression of the foot: $tan 2 6^{\circ} = \frac{18}{d}$ , so $d = \frac{18}{0.4877} \approx 36.91$ m.
Horizontal from the building top meets the tower at 18 m. Elevation: extra height $= d tan 3 4^{\circ} = 36.908 \times 0.6745 \approx 24.89$ m.
Tower $= 18 + 24.894 \approx 42.89$ m.

Answer: Gap ≈ 36.91 m; tower ≈ 42.89 m

Example 4: Depressions of the top and foot of a shorter tower · 5 marks · Long answerHOTS / competency

From the top of a building 60 m high, the angles of depression of the top and the foot of a shorter tower on the same level ground are $2 3^{\circ}$ and $4 7^{\circ}$ . Find the distance between them and the height of the tower, to 2 decimal places. ( $tan 2 3^{\circ} = 0.4245$ , $tan 4 7^{\circ} = 1.0724$ )

Try it first, then show the solutionSolution

Foot: $tan 4 7^{\circ} = \frac{60}{d}$ , so $d = \frac{60}{1.0724} \approx 55.95$ m.
Draw the horizontal through the tower top to meet the building at $P$ . The drop from the building top to $P$ $= d tan 2 3^{\circ} = 55.949 \times 0.4245 \approx 23.75$ m.
Tower $= 60 - 23.751 \approx 36.25$ m.

Answer: Gap ≈ 55.95 m; tower ≈ 36.25 m

Example 5: Flagstaff on a tower · 3 marks · Short answerHarder board level

From a point 40 m from the foot of a tower on level ground, the angles of elevation of the top of the tower and of the top of a flagstaff fixed on it are $3 7^{\circ}$ and $4 6^{\circ}$ . Find the height of the tower and the length of the flagstaff, to 2 decimal places. ( $tan 3 7^{\circ} = 0.7536$ , $tan 4 6^{\circ} = 1.0355$ )

Try it first, then show the solutionSolution

Tower $= 40 tan 3 7^{\circ} = 40 \times 0.7536 \approx 30.14$ m.
Top of flagstaff $= 40 tan 4 6^{\circ} = 41.42$ m.
Flagstaff $= 40 (1.0355 - 0.7536) = 40 \times 0.2819 \approx 11.28$ m.

Answer: Tower ≈ 30.14 m; flagstaff ≈ 11.28 m

Example 6: Case study: from the roof and from half-way up · 5 marks · Case-basedHOTS / competency

A tower $A B$ stands 80 m from a building $C D$ on level ground ( $B$ and $D$ are the feet). $E$ is the mid-point of $C D$ , and $P$ is the point on $C D$ level with $A$ . The angles of depression of $A$ from $C$ and from $E$ are $4 0^{\circ}$ and $1 5^{\circ}$ . ( $tan 4 0^{\circ} = 0.8391$ , $tan 1 5^{\circ} = 0.2679$ )
(a) Find $C P$ and $E P$ , to 2 decimal places.
(b) Find the heights of the building and the tower, to 1 decimal place.
(c) Find, to the nearest degree, the angle of elevation of $C$ from $B$ . ( $tan 4 8^{\circ} = 1.1106$ , $tan 4 9^{\circ} = 1.1504$ )

Try it first, then show the solutionSolution

(a) $A P = B D = 80$ m. $C P = 80 tan 4 0^{\circ} = 67.128 \approx 67.13$ m; $E P = 80 tan 1 5^{\circ} = 21.432 \approx 21.43$ m.
(b) $C E = C P - E P = 45.696$ m. $E$ is the mid-point, so $C D = 2 \times 45.696 \approx 91.4$ m. $A B = P D = E D - E P = 45.696 - 21.432 \approx 24.3$ m.
(c) $tan θ = \frac{C D}{B D} = \frac{91.392}{80} \approx 1.14$ . This lies between $tan 4 8^{\circ}$ and $tan 4 9^{\circ}$ and is nearer 1.1504, so $θ \approx 4 9^{\circ}$ .

Answer: (a) 67.13 m, 21.43 m (b) building ≈ 91.4 m, tower ≈ 24.3 m (c) ≈ $4 9^{\circ}$

Where students lose marks in Two Vertical Objects

The slips examiners see most often, and what to write instead.

Mistake: No horizontal line from the observer
Without it there is no right triangle for the upper or lower part of the other object.
Wrong working
Using the full tower height with the elevation angle from the building top
Right working
Elevation gives only the part above the observer's level: $d tan α$
Do this: Draw the dashed horizontal across and label the point where it meets the other object.
Usually costs: The CISCE examiners' suggestions for 2025 Q6(ii) ask for labelled diagrams with depression marked from the horizontal.
Mistake: Inverted tan ratio for the vertical piece
With the gap known, the vertical piece is opposite and the gap adjacent.
Wrong working
$tan 1 4^{\circ} = \frac{100}{E P}$
Right working
$tan 1 4^{\circ} = \frac{E P}{100}$ , so $E P = 100 tan 1 4^{\circ}$
Do this: Label opposite and adjacent before writing any ratio.
Usually costs: Reported in the CISCE examiners' comments on 2025 Q6(ii); the building height then came out wrong.
Mistake: Adding when you should subtract
A depression of the other top means that top is below your eye.
Wrong working
Tower $= H + d tan α$ when $α$ is a depression
Right working
Tower $= H - d tan α$
Do this: Ask: is the other top above or below my horizontal?
Usually costs: The final answer mark.
Mistake: Rounding the gap before the second triangle
The rounding error is multiplied by $tan α$ .
Wrong working
$d \approx 37$ m, then $37 \times 0.6745$
Right working
$d \approx 36.908$ m, then $36.908 \times 0.6745$
Do this: Carry three or four decimals; round only the final height.
Usually costs: Reported in the CISCE examiners' comments on 2025 Q6(ii): rounding during the working.

Exam tips

Asked in 2025 Q6(ii) (5 marks): a tower 100 m from a 30-storey building, depressions of $3 5^{\circ}$ and $1 4^{\circ}$ from the top and mid-point; the expected route was $E P = 100 tan 1 4^{\circ}$ , $C P = 100 tan 3 5^{\circ}$ , $C E = C P - E P$ , then $C D = 2 C E$ and $A B = E D - E P$ .
When the question gives a rounding rule for table values (2025 Q6(ii): two decimals), use it exactly, then follow it to the end.
Name every point you add (such as $P$ , where the horizontal meets the other object) and use it in your equations.
Check the answer: a tower seen with an angle of depression must come out shorter than the building.

Practice questions on Two Vertical Objects

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Two towers are 30 m and 50 m high. From the top of the shorter one, the angle of elevation of the top of the taller one is $4 5^{\circ}$ . The distance between the towers is:
(a) 20 m
(b) 30 m
(c) 50 m
(d) 80 m
Hint
The difference in heights is opposite the $4 5^{\circ}$ angle.
Answer
(a)
Q2 · 3 marks · Short answerBoard level
From a point on the ground midway between two poles 60 m apart, the angles of elevation of their tops are $3 0^{\circ}$ and $6 0^{\circ}$ . Find the heights of the poles, to 2 decimal places.
Hint
Each pole is 30 m from the point.
Answer
$30 tan 3 0^{\circ} = 103 \approx 17.32$ m and $30 tan 6 0^{\circ} = 303 \approx 51.96$ m
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): From the top of a building 20 m high, the angle of elevation of the top of a tower equals the angle of depression of its foot. Then the tower is 40 m high.
Reason (R): When the two angles are equal, the parts of the tower above and below the observer's horizontal are equal.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Both parts equal $d tan θ$ .
Answer
(c)
Q4 · 4 marks · Short answerHarder board level
From the top of a building 25 m high, the angles of depression of the top and the foot of a lamp post are $2 1^{\circ}$ and $3 2^{\circ}$ . Find the height of the lamp post, to 2 decimal places. ( $tan 2 1^{\circ} = 0.3839$ , $tan 3 2^{\circ} = 0.6249$ )
Hint
Gap from the foot first, then subtract the drop.
Answer
$d = \frac{25}{0.6249} \approx 40.01$ m; drop $= 40.006 \times 0.3839 \approx 15.36$ m; lamp post $\approx 25 - 15.358 \approx 9.64$ m.
Q5 · 3 marks · Short answerHarder board level
The angle of elevation of the top of a building 36 m high from a point on the ground is $3 3^{\circ}$ . From the same point, the angle of elevation of the top of a mast on the building is $4 4^{\circ}$ . Find the length of the mast, to 1 decimal place. ( $tan 3 3^{\circ} = 0.6494$ , $tan 4 4^{\circ} = 0.9657$ )
Hint
Find the distance from the building first.
Answer
$d = \frac{36}{0.6494} \approx 55.44$ m; mast top $\approx 55.436 \times 0.9657 \approx 53.53$ m; mast $\approx 17.5$ m.
Q6 · 3 marks · Short answerHarder board level
Two towers $A$ and $B$ stand 50 m apart on level ground. From the foot of $A$ , the angle of elevation of the top of $B$ is $5 5^{\circ}$ ; from the foot of $B$ , the angle of elevation of the top of $A$ is $3 2^{\circ}$ . Find both heights, to 2 decimal places. ( $tan 5 5^{\circ} = 1.4281$ , $tan 3 2^{\circ} = 0.6249$ )
Hint
Two separate triangles with the same base.
Answer
$B = 50 \times 1.4281 \approx 71.41$ m; $A = 50 \times 0.6249 \approx 31.25$ m.
Q7 · 4 marks · Short answerHOTS / competency
From the top of a building of height $h$ , the angle of elevation of the top of a tower is $α$ and the angle of depression of its foot is $β$ . Prove that the height of the tower is $h (1 + \frac{t a n α}{t a n β})$ , and find it when $h = 12$ m, $α = 3 0^{\circ}$ and $β = 6 0^{\circ}$ .
Hint
$d = \frac{h}{t a n β}$ , extra $= d tan α$ .
Answer
Tower $= h + \frac{h}{t a n β} tan α = h (1 + \frac{t a n α}{t a n β})$ . With the values: $12 (1 + \frac{1/ 3}{3}) = 12 \times \frac{4}{3} = 16$ m.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.From the top of a building, which angle gives the gap?Show answer
The depression of the other object's foot: $d = H cot β$ .
2.Taller tower seen from the building top: height?Show answer
$H + d tan α$
3.Shorter tower (depression of its top): height?Show answer
$H - d tan α$
4.Flagstaff on a tower, seen from distance $d$ : length?Show answer
$d (tan β - tan α)$
5.Equal elevation and depression from a building of height $H$ : tower?Show answer
$2 H$

Two Vertical Objects: frequently asked questions

How do I solve problems with a building and a tower?

Draw a horizontal line from the observer at the top of the building across to the tower. Use the depression of the foot to find the gap, then the elevation (or depression) of the tower top to find the part above (or below) that line.

What if the gap between the two objects is given?

Then each angle gives a vertical length at once: length = gap × tan(angle). Combine these lengths along the building or tower, as in 2025 Q6(ii).

Are three-triangle problems in the ICSE syllabus?

No. The CISCE syllabus excludes cases involving more than two right-angled triangles, so building-and-tower questions are set with two triangles.