MathsCBSE Class 10 · Some Applications of Trigonometry

Heights and Distances: Problems with One Right Triangle

Updated 13 October 2026 · Fact-checked

A one right triangle heights and distances problem uses a single known length and one angle of 30, 45 or 60 degrees to find a height, a ground distance or a slanted length. The board asks 1-mark MCQs, 2- and 3-mark short answers, and multi-step long answers and case studies built on one triangle, often with an observer's eye height.

Find the vertical, horizontal and slant sides from angle θ, then choose tan, sin or cos by the two sides involved.

Illustration drawn for QuestionBuddy.

Understand Heights and Distances: One Right Triangle

Every heights and distances problem in this topic hides one right triangle. The vertical object (tower, tree, pole, wall) stands at a right angle to level ground. The line of sight, or a rope, ladder or kite string, is the slanted side. So you always have a vertical side, a horizontal side and a hypotenuse, with the right angle at the foot of the object.

Why does one angle and one length solve the whole triangle? In a right triangle, once one acute angle is fixed, the shape is fixed. All triangles with that angle are similar, so the ratio of any two sides is always the same number. That number is exactly what $tan$ , $sin$ and $cos$ give you. Multiply the known side by the right ratio and you get the unknown side.

The skill is choosing the ratio. Name the side you know and the side you want, relative to the given angle:

height and ground distance: $tan θ = \frac{height}{distance}$
height and slanted length: $sin θ = \frac{height}{slant}$
ground distance and slanted length: $cos θ = \frac{distance}{slant}$

You can also run it backwards. If two sides are known, form the ratio and match it with a standard value to find the angle, for example $tan θ = 3$ gives $θ = 6 0^{\circ}$ .

Board questions turn one triangle into a 3- or 5-mark problem in a few ways: an observer's eye height that you must subtract first, a broken tree where the two pieces add to the original height, a condition such as 'the rope is 6 m longer than the distance' that sets up an equation, or a cost or speed calculated from your answer. The triangle is still one triangle; only the modelling is harder.

The method holds only when the ground is level, the object is vertical, and the angle is measured from the horizontal. The syllabus keeps angles to $3 0^{\circ}$ , $4 5^{\circ}$ and $6 0^{\circ}$ , so every answer comes out in terms of $1$ , $2$ or $3$ . Problems with two triangles (an observer who moves, or two objects) build on exactly this skill and are covered on their own page.

Heights and Distances: One Right Triangle: formulas to remember

Values at 30°: $sin 3 0^{\circ} = \frac{1}{2}, cos 3 0^{\circ} = \frac{3}{2}, tan 3 0^{\circ} = \frac{1}{3}$; The side opposite 30° is half the hypotenuse.
Values at 45°: $sin 4 5^{\circ} = cos 4 5^{\circ} = \frac{1}{2}, tan 4 5^{\circ} = 1$; Height equals ground distance; the slant is √2 times either side.
Values at 60°: $sin 6 0^{\circ} = \frac{3}{2}, cos 6 0^{\circ} = \frac{1}{2}, tan 6 0^{\circ} = 3$; The ground side next to 60° is half the hypotenuse.
Height from ground distance: $h = d tan θ$; Use when the horizontal distance d is known and the vertical height h is wanted.
Ground distance from height: $d = \frac{h}{t a n θ}$; At 30° this is h√3, at 45° it is h, at 60° it is h/√3.
Height from slanted length: $h = L sin θ$; L is a rope, ladder, kite string or line of sight (the hypotenuse).
Ground distance from slanted length: $d = L cos θ$; Use for the foot of a ladder or the point where a rope meets the ground.
Sun's altitude and shadow: $tan (altitude) = \frac{height}{shadow}$; The altitude of the sun is the angle of elevation of the sun from the tip of the shadow.
Observer with eye height p: $H = p + d tan θ$; Solve the triangle above eye level first, then add the eye height.
Finding the angle: $tan θ = \frac{h}{d} : \frac{1}{3} \to 3 0^{\circ}, 1 \to 4 5^{\circ}, 3 \to 6 0^{\circ}$; Simplify the ratio fully before matching it with a standard value. With sin or cos, match against the value rows above.
Broken tree: $original height = standing part + broken part$; The broken part is the hypotenuse; it lies from the break point to the ground.
Pythagoras check: $L^{2} = h^{2} + d^{2}$; Quick check on any answer that has all three sides.

How to solve Heights and Distances: One Right Triangle questions

Use this routine on every one-triangle question. It keeps your figure, ratio and answer correct and shows the examiner each step.

1Draw level ground as a horizontal line and the object as a vertical line; mark the right angle at its foot.
2If an observer's eye height is given, draw the horizontal from the eye and note that the triangle's vertical side is (object height minus eye height).
3Mark the given angle between the horizontal and the slanted line, at the correct point.
4Label the known side with its value and the unknown with h, d or L. If a condition links two sides, write it as an equation.
5Pick the ratio containing the known and unknown sides: tan for height and distance, sin for height and slant, cos for distance and slant.
6Substitute the standard value, solve, and rationalise any surd in the denominator.
7Add back the eye height or other parts if needed, then convert to decimals only if the question gives the value of √3.
8Check with Pythagoras or the side ratio 1 : √3 : 2, and write the final answer with units.

Heights and Distances: One Right Triangle 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.

Heights and Distances: One Right Triangle

Suppose on a festival afternoon you fly a kite from your terrace in an open area. You hold the string 1.2 m above the terrace floor. At first 80 m of string is out, taut, making an angle of $3 0^{\circ}$ with the horizontal. Treat the string as a straight line, and imagine extra string is sold in reels of 25 m at ₹30 per reel.

Find the kite's height above the terrace floor at first. You want the kite to reach 61.2 m above the floor with the string still at $3 0^{\circ}$ . How much extra string must you let out, how many reels must you buy, and what do they cost?

Try it first, then show the solutionSolution

The string is the hypotenuse and the height above your hand is opposite the $3 0^{\circ}$ angle, so use sin.
At first: height above hand $= 80 sin 3 0^{\circ} = 80 \times \frac{1}{2} = 40$ m.
Height above floor $= 40 + 1.2 = 41.2$ m.
Target height above hand $= 61.2 - 1.2 = 60$ m.
String needed: $sin 3 0^{\circ} = \frac{60}{L}$ , so $L = \frac{60}{1/2} = 120$ m.
Extra string $= 120 - 80 = 40$ m.
One reel gives only 25 m, which is not enough, so you need 2 reels (50 m), leaving 10 m spare.
Cost $= 2 \times 30 =$ ₹60.

Answer: At first the kite is 41.2 m above the terrace floor. You must let out 40 m more string, so you buy 2 reels costing ₹60.

Memory hook: Kite string is the slant: take off your hand's height, then divide by sin to find the string you need.

Tricks to remember Heights and Distances: One Right Triangle

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

T-H-D, S-H-L, C-D-L
Tan links Height and Distance. Sin links Height and Length (slant). Cos links Distance and Length.
Use it when: Say it before writing any ratio so you pick the right one first time.
Shortest side faces 30°
In a 30°-60°-90° triangle the sides are 1 : √3 : 2; the side opposite 30° is the smallest, the hypotenuse is double it.
Use it when: Check an answer in seconds: if the height at 30° comes out larger than the distance, your ratio is upside down.
Eye first, add last
Subtract the eye height before using tan, add it back at the end.
Use it when: Any question that says 'a man 1.5 m tall' or 'his eyes are at height'.
45 means equal
At 45°, height equals ground distance.
Use it when: Spot quick MCQ answers and check your figure.

Heights and Distances: One Right Triangle: solved examples

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

Example 1: Broken tree: find the original height · 2 marks · Very short answerBoard level

A vertical tree is broken by a storm. The broken part bends so that its top touches the level ground at a point 8√3 m from the foot of the tree, making an angle of $3 0^{\circ}$ with the ground. The broken part is still attached to the standing trunk. Find the height of the tree before it broke.

Try it first, then show the solutionSolution

Let the trunk stand as AB (A on the ground), broken at B, with the top touching the ground at C; AC = $83$ m, $∠ A C B = 3 0^{\circ}$ , $∠ B A C = 9 0^{\circ}$ .
In $△ B A C$ , $cos 3 0^{\circ} = \frac{A C}{B C}$ , so $B C = \frac{8 3}{3 /2} = 16$ m. (1 mark)
$tan 3 0^{\circ} = \frac{A B}{A C}$ , so $A B = \frac{8 3}{3} = 8$ m. (½ mark)
Original height $= A B + B C = 8 + 16 = 24$ m. (½ mark)
Check: $8^{2} + (83)^{2} = 64 + 192 = 256 = 1 6^{2}$ .

Answer: The tree was 24 m tall.

Example 2: Observer's eye height and line of sight · 3 marks · Short answerBoard level

A boy whose eyes are 1.5 m above the ground stands on level ground at a horizontal distance of $153$ m from a vertical building. The angle of elevation of the top of the building from his eyes is $6 0^{\circ}$ . Find (i) the height of the building and (ii) the length of his line of sight to the top. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Let the building be PQ (Q on the ground) and the eye be at E, with the horizontal from E meeting PQ at R; QR = 1.5 m, ER = $153$ m, $∠ P E R = 6 0^{\circ}$ .
In right $△ P R E$ , $tan 6 0^{\circ} = \frac{P R}{E R}$ , so $P R = 153 \times 3 = 45$ m. (1 mark)
Height $P Q = P R + R Q = 45 + 1.5 = 46.5$ m. (½ mark)
$sin 6 0^{\circ} = \frac{P R}{P E}$ , so $P E = \frac{45}{3 /2} = \frac{90}{3} = 303$ m. (1 mark)
$P E = 30 \times 1.732 = 51.96$ m. (½ mark)

Answer: (i) 46.5 m (ii) $303$ m, about 51.96 m.

Example 3: Find the angle, then a distance · 3 marks · Short answerBoard level

A kite is flying at a height of 60 m above level ground. Its string is tied to a peg on the ground and is taut, with length $403$ m. Treating the string as straight, find the angle the string makes with the ground and the horizontal distance of the peg from the point directly below the kite. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Let the kite be K, the point below it on the ground M and the peg P; KM = 60 m, KP = $403$ m, $∠ K M P = 9 0^{\circ}$ .
$sin θ = \frac{K M}{K P} = \frac{60}{40 3} = \frac{3}{2 3} = \frac{3}{2}$ , so $θ = 6 0^{\circ}$ . (1½ marks)
$P M = K P cos 6 0^{\circ} = 403 \times \frac{1}{2} = 203 = 34.64$ m. (1 mark)
Check: $6 0^{2} + (203)^{2} = 3600 + 1200 = 4800 = (403)^{2}$ . (½ mark)

Answer: The string makes $6 0^{\circ}$ with the ground; the peg is $203$ m, about 34.64 m, from the point below the kite.

Example 4: Stay wire: set up an equation, then find cost · 5 marks · Long answerHarder board level

A straight wire is stretched from the top of a vertical pole to a peg on level ground. The wire makes an angle of $6 0^{\circ}$ with the ground, and its length is 6 m more than the distance of the peg from the foot of the pole. (i) Find the distance of the peg from the foot, the length of the wire and the height of the pole. (ii) If the wire costs ₹40 per metre, find the cost of the wire. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Let the pole be AB (B on the ground) and the peg C; let BC = d m and AC = L m, with $∠ A C B = 6 0^{\circ}$ and $∠ A B C = 9 0^{\circ}$ .
$cos 6 0^{\circ} = \frac{B C}{A C} = \frac{d}{L}$ , so $L = 2 d$ . (1 mark)
Given $L = d + 6$ , so $2 d = d + 6$ , giving $d = 6$ m. (1 mark)
$L = 2 \times 6 = 12$ m. (½ mark)
$tan 6 0^{\circ} = \frac{A B}{B C}$ , so $A B = 63 = 10.39$ m. (1 mark)
Check: $6^{2} + (63)^{2} = 36 + 108 = 144 = 1 2^{2}$ . (½ mark)
Cost $= 12 \times 40 =$ ₹480. (1 mark)

Answer: (i) Peg 6 m from the foot, wire 12 m, pole $63$ m (about 10.39 m). (ii) ₹480.

Example 5: Case study: fire-brigade ladder · 4 marks · Case-basedHOTS / competency

A fire-brigade truck has a ladder 20 m long whose lower end is fixed on the truck 2 m above level ground. The ladder can be raised to any angle with the horizontal and its top rests against a vertical building wall.
(i) At $6 0^{\circ}$ with the horizontal, how high up the wall does the ladder reach? (1 mark)
(ii) At $6 0^{\circ}$ , what is the horizontal distance from the ladder's lower end to the wall? (1 mark)
(iii) A trapped person is at a window 12 m above the ground. At what angle must the full ladder be set, and how far from the wall must its lower end be? (2 marks) Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Height above the lower end $= 20 sin 6 0^{\circ} = 103 = 17.32$ m, so height on wall $= 2 + 17.32 = 19.32$ m. (1 mark)
Horizontal distance $= 20 cos 6 0^{\circ} = 10$ m. (1 mark)
Vertical rise needed $= 12 - 2 = 10$ m, so $sin θ = \frac{10}{20} = \frac{1}{2}$ , giving $θ = 3 0^{\circ}$ . (1 mark)
Horizontal distance $= 20 cos 3 0^{\circ} = 103 = 17.32$ m. (1 mark)

Answer: (i) about 19.32 m (ii) 10 m (iii) $3 0^{\circ}$ , with the lower end $103$ m, about 17.32 m, from the wall.

Where students lose marks in Heights and Distances: One Right Triangle

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

Mistake: Writing tan upside down
The student forgets which side is opposite the angle at the observer.
Wrong working
Distance 20 m, elevation $3 0^{\circ}$ : $tan 3 0^{\circ} = \frac{20}{h}$ , so $h = 203$ m.
Right working
$tan 3 0^{\circ} = \frac{h}{20}$ , so $h = \frac{20}{3} = \frac{20 3}{3}$ m.
Do this: Height is always opposite the angle at the ground point. Check: at 30°, height must be less than distance.
Usually costs: Most of the marks for the question.
Mistake: Ignoring the observer's eye height
The triangle is drawn from the ground instead of from eye level.
Wrong working
Eye 1.5 m, distance $153$ m, elevation $6 0^{\circ}$ : height $= 45$ m.
Right working
$45$ m is the part above eye level; height $= 45 + 1.5 = 46.5$ m.
Do this: Draw the horizontal from the eye and write 'height = part above eye + eye height'.
Usually costs: The final ½ to 1 mark, plus any later part that uses it.
Mistake: Using tan when the slanted side is given
Tan is treated as the default ratio for every question.
Wrong working
Ladder 10 m at $6 0^{\circ}$ : $tan 6 0^{\circ} = \frac{h}{10}$ , so $h = 103$ m.
Right working
The ladder is the hypotenuse: $h = 10 sin 6 0^{\circ} = 53$ m.
Do this: If a rope, ladder, string or line of sight is known, use sin or cos.
Usually costs: Almost all the marks for that part.
Mistake: Rounding √3 too early
Decimals are substituted in the first line, and later steps compound the error.
Wrong working
$B C = \frac{8 \times 1.73}{0.87} \approx 15.9$ m, so the tree is about 23.9 m.
Right working
$B C = \frac{8 3}{3 /2} = 16$ m exactly, so the tree is 24 m.
Do this: Keep surds until the last line; convert only if √3 = 1.732 is given.
Usually costs: Accuracy mark, and a wrong value in MCQs.
Mistake: Measuring the angle from the vertical
The angle is marked at the top of the object instead of at the observer.
Wrong working
Elevation $3 0^{\circ}$ , distance 30 m, angle placed at the top: $h = 30 tan 6 0^{\circ} = 303$ m.
Right working
Angle at the observer: $h = 30 tan 3 0^{\circ} = 103$ m.
Do this: Elevation and depression are always measured from the horizontal through the eye.
Usually costs: The whole question, since the figure is wrong.

Exam tips

Draw a neat, labelled figure in every written answer, with the right angle and the given angle marked; examiners expect it.
Write the ratio in words or letters first, such as tan 60° = AB/BC, then substitute. This earns method marks even if arithmetic slips.
Leave answers in surd form unless the question gives √3 = 1.732; when it does, convert only in the final line.
In case studies, the 1-mark parts often feed the 2-mark part, so keep exact values and reuse them.
Finish with a quick check: Pythagoras, the 1 : √3 : 2 ratio, or whether the height should be more or less than the distance.

Practice questions on Heights and Distances: One Right Triangle

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
A vertical tower 15 m tall casts a shadow of length $53$ m on level ground. The altitude of the sun is:
(a) $3 0^{\circ}$
(b) $4 5^{\circ}$
(c) $6 0^{\circ}$
(d) $9 0^{\circ}$
Hint
Form tan of the angle as height over shadow and simplify.
Answer
(c) $6 0^{\circ}$ . $tan θ = \frac{15}{5 3} = 3$ .
Q2 · 5 marks · Long answerHarder board level
A ladder rests against a vertical wall on level ground, making an angle of $6 0^{\circ}$ with the ground. The height at which its top touches the wall is $2 (3 - 1)$ m more than the distance of its foot from the wall. (i) Find the distance of the foot from the wall, the length of the ladder and the height it reaches. (ii) The foot is then pulled out so that the ladder makes $3 0^{\circ}$ with the ground, its top still on the wall. By how much does the top slide down? Use $3 = 1.732$ .
Hint
Let the foot distance be d. At 60°, height = d√3 and ladder = 2d; use the condition to form an equation in d.
Answer
(i) $tan 6 0^{\circ} = \frac{h}{d}$ gives $h = d 3$ , so $d 3 - d = 2 (3 - 1)$ , giving $d = 2$ m. Ladder $= \frac{d}{c o s 6 0 ^{\circ}} = 4$ m, height $= 23 \approx 3.46$ m. Check: $4 + 12 = 16 = 4^{2}$ . (ii) At $3 0^{\circ}$ , height $= 4 sin 3 0^{\circ} = 2$ m, so the top slides down $23 - 2 = 2 (3 - 1) \approx 1.46$ m.
Q3 · 4 marks · Case-basedHOTS / competency
Case study: Your school puts up a vertical flagpole on level ground. A straight rope is tied from the top of the pole to a peg on the ground 5 m from the foot of the pole, and the rope makes an angle of $6 0^{\circ}$ with the ground.
(i) Find the height of the flagpole. (1 mark)
(ii) Find the length of the rope. (1 mark)
(iii) The peg is moved further out along the same line so that a new rope from the top makes $3 0^{\circ}$ with the ground. Find the length of the new rope and the cost of the extra rope needed over the old length, at ₹25 per metre. (2 marks) Use $3 = 1.732$ .
Hint
Use tan 60° for the height and cos 60° for the rope, then sin 30° with the same height in part (iii).
Answer
(i) $h = 5 tan 6 0^{\circ} = 53 \approx 8.66$ m. (ii) Rope $= \frac{5}{c o s 6 0 ^{\circ}} = 10$ m. (iii) New rope $= \frac{5 3}{s i n 3 0 ^{\circ}} = 103 \approx 17.32$ m (1 mark). Extra rope $= 17.32 - 10 = 7.32$ m, cost $= 7.32 \times 25 =$ ₹183 (1 mark).
Q4 · 5 marks · Long answerHarder board level
A vertical tree 18 m tall breaks at a point, and the broken part (still attached) bends so its top touches the level ground, making an angle of $3 0^{\circ}$ with the ground. Find the height at which the tree broke, the length of the broken part, and the distance from the foot of the tree to the point where the top touches the ground. Use $3 = 1.732$ .
Hint
Let the standing part be h, so the broken part (hypotenuse) is 18 - h. Use sin 30°.
Answer
$sin 3 0^{\circ} = \frac{h}{18 - h}$ gives $2 h = 18 - h$ , so $h = 6$ m. Broken part $= 12$ m. Distance $= 12 cos 3 0^{\circ} = 63 \approx 10.39$ m. Check: $36 + 108 = 144 = 1 2^{2}$ .
Q5 · 3 marks · Short answerHOTS / competency
A point P on level ground is 10 m from the foot of a vertical pole, and the angle of elevation of the top of the pole from P is $6 0^{\circ}$ . A student wrote: $tan 6 0^{\circ} = \frac{10}{h}$ , so $h = \frac{10}{3} \approx 5.77$ m. (i) Explain the error, using a quick reasonableness check. (ii) Find the correct height. Use $3 = 1.732$ .
Hint
Tan is opposite over adjacent with respect to the angle at P. At 60° should the height be more or less than the distance?
Answer
(i) The student put the adjacent side (10 m) over the opposite side (h). At $6 0^{\circ}$ , tan is greater than 1, so the height must exceed 10 m; 5.77 m cannot be right. (ii) $tan 6 0^{\circ} = \frac{h}{10}$ , so $h = 103 \approx 17.32$ 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.Which ratio links height and ground distance?Show answer
$tan θ = \frac{height}{distance}$
2.A ladder of length L makes angle θ with the ground. How high does it reach?Show answer
$L sin θ$
3.Value of tan 60°?Show answer
$3$
4.At 45°, how do height and distance compare?Show answer
They are equal.
5.Sides of a 30°-60°-90° triangle?Show answer
$1 : 3 : 2$ , shortest side opposite $3 0^{\circ}$ .
6.Eye height p, distance d, elevation θ: object height?Show answer
$p + d tan θ$

Heights and Distances: One Right Triangle: frequently asked questions

How do I know whether to use sin, cos or tan?

Name the side you know and the side you want, relative to the given angle. Height and ground distance need tan; if a slanted length such as a ladder or string is involved, use sin with the height or cos with the ground distance.

Is a figure compulsory in heights and distances answers?

A clear, labelled figure is expected in written answers and helps the examiner follow your working. It also stops most mistakes, so always draw one.

Should I give the answer in surd form or decimals?

Give the surd form unless the question gives a value like 1.732 for root 3 or asks for a decimal. If a value is given, convert only at the end.

Can the board ask angles other than 30, 45 and 60 degrees?

No. The 2026-27 syllabus limits angles of elevation and depression to 30, 45 and 60 degrees, and problems to at most two right triangles.

Does Exercise 9.1 only have one-triangle questions?

No. Exercise 9.1 mixes one-triangle questions, such as ladders, strings and broken trees, with two-triangle questions where an observer moves or looks at two objects. Master the one-triangle method first, since every two-triangle problem is built from it.