MathsCBSE Class 10 · Some Applications of Trigonometry

Line of Sight and Angle of Elevation for Class 10

Updated 14 October 2026 · Fact-checked

The line of sight is the straight line from your eye to an object. When the object is above your eye level, the angle between the horizontal through your eye and the line of sight is the angle of elevation. The board asks you to draw this right triangle and use tan, sin or cos at 30, 45 or 60°.

The angle of elevation θ is measured up from the horizontal at your eye; height is opposite θ, ground distance is adjacent, so use tan θ.

Illustration drawn for QuestionBuddy.

Understand Line of Sight and Angle of Elevation

Stand on level ground and look at the top of a tower. The straight line from your eye to the top is the line of sight. The flat line from your eye parallel to the ground is the horizontal level. Because the top is above your eye, you tilt your head up. The angle between the horizontal and the line of sight is the angle of elevation.

Why does this help? A vertical tower meets level ground at $9 0^{\circ}$ . So the ground distance, the tower and the line of sight form a right triangle, with the right angle at the foot and the angle of elevation at your eye. The height is the side opposite the angle, the ground distance is the adjacent side, and the line of sight is the hypotenuse.

Now every ratio from Chapter 8 applies. Height and ground distance are linked by $tan θ = \frac{height}{distance}$ . If the slanted length (a kite string, a ladder, the line of sight itself) is involved, use $sin θ$ or $cos θ$ . Know one side and the angle, and you can find the rest.

The angle is always measured from the horizontal, never from the vertical object. If you are given the angle between the line of sight and the tower, the elevation is $9 0^{\circ}$ minus it. Also, the horizontal starts at your eye. If your eyes are $1.5$ m above the ground, the triangle's vertical side is the part of the object above eye level, so you add $1.5$ m at the end.

The angle tells you shape. At $4 5^{\circ}$ the height equals the distance. At $3 0^{\circ}$ the height is smaller than the distance; at $6 0^{\circ}$ it is larger. As you walk closer, the angle of elevation increases. This is the base for the other topics: angle of depression, two-triangle problems and word problems.

Conditions for 2026-27: ground is level, objects are vertical, angles are only $3 0^{\circ}$ , $4 5^{\circ}$ or $6 0^{\circ}$ , and no problem uses more than two right triangles.

Line of Sight and Angle of Elevation: formulas to remember

Height and ground distance: $tan θ = \frac{height}{horizontal distance}$; Most used. Gives height = distance × tan θ.
Height and line of sight: $sin θ = \frac{height}{line of sight}$; For kite strings, ropes, ladders.
Distance and line of sight: $cos θ = \frac{horizontal distance}{line of sight}$; Use when the slant length (string, ladder, line of sight) and the ground distance are involved.
Observer with eye height: $H = p + d tan θ$; p is eye height, d horizontal distance.
Standard tan values: $tan 3 0^{\circ} = \frac{1}{3}, tan 4 5^{\circ} = 1, tan 6 0^{\circ} = 3$; Match a ratio to find the angle.
Standard sin and cos values: $sin 3 0^{\circ} = \frac{1}{2}, sin 4 5^{\circ} = \frac{1}{2}, sin 6 0^{\circ} = \frac{3}{2}; cos 3 0^{\circ} = \frac{3}{2}, cos 4 5^{\circ} = \frac{1}{2}, cos 6 0^{\circ} = \frac{1}{2}$; Use these with a slant length (ladder, string, line of sight). When the observer moves, the distance moved is the difference of the two horizontal distances.
Angle given from the vertical: $elevation = 9 0^{\circ} - (angle with vertical)$; Elevation is always from the horizontal.

How to solve Line of Sight and Angle of Elevation questions

Use this routine on any angle of elevation question.

1Draw level ground and the object as a vertical line; mark the right angle at its foot.
2Mark the observer's eye. If the observer has height, draw the horizontal at eye level.
3Mark the angle of elevation between the horizontal and the line of sight.
4Label known lengths; call unknowns h and x.
5Choose the ratio containing the known side and the wanted side (tan, sin or cos).
6Substitute the standard value, solve, and rationalise any surd.
7Add the eye height if needed, write units, and check with the 1 : √3 : 2 or 1 : 1 : √2 ratio.

Line of Sight and Angle of Elevation 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.

Line of sight and angle of elevation

Imagine your school hosts a kite-flying contest on a level ground. Your kite's string is 60 m long, held taut and straight from your hand, which is 1.2 m above the ground. The string makes an angle of $6 0^{\circ}$ with the horizontal at your hand. Your friend's kite string is 40 m long at $3 0^{\circ}$ , held at the same hand height.

Find the height of each kite above the ground and how much higher your kite is. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

The string is the line of sight (hypotenuse); use $sin$ .
Your kite above hand: $60 sin 6 0^{\circ} = 60 \times \frac{3}{2} = 303 = 51.96$ m.
Your kite height: $51.96 + 1.2 = 53.16$ m.
Friend's kite above hand: $40 sin 3 0^{\circ} = 20$ m, height $21.2$ m.
Difference: $53.16 - 21.2 = 31.96$ m.

Answer: Your kite 53.16 m, friend's 21.2 m; yours is 31.96 m higher.

Memory hook: Kite string is the slant: string means sine.

Line of sight and angle of elevation

Suppose your family visits a hill town and you stand on a level road 50 m from the foot of a vertical mobile tower. Your eyes are 1.6 m above the road. The angle of elevation of the top of the tower from your eyes is $4 5^{\circ}$ , and of a platform on the tower it is $3 0^{\circ}$ .

Find the height of the tower and the height of the platform above the road. Then find how far you must walk towards the tower for the angle of elevation of its top to become $6 0^{\circ}$ . Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Top above eye level: $50 tan 4 5^{\circ} = 50$ m.
Tower height: $50 + 1.6 = 51.6$ m.
Platform above eye level: $50 tan 3 0^{\circ} = \frac{50}{3} = \frac{50 3}{3} \approx 28.87$ m.
Platform height: $28.87 + 1.6 = 30.47$ m.
For an elevation of $6 0^{\circ}$ to the top (still 50 m above your eyes), the distance must be $\frac{50}{t a n 6 0 ^{\circ}} = \frac{50}{3} \approx 28.87$ m.
So walk $50 - 28.87 = 21.13$ m towards the tower.

Answer: Tower 51.6 m, platform about 30.47 m above the road; walk about 21.13 m closer for an elevation of $6 0^{\circ}$ .

Memory hook: Eyes are the starting line: solve above them, then add your eye height.

Tricks to remember Line of Sight and Angle of Elevation

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

Up from flat
Elevation is measured UP FROM the FLAT horizontal, never from the pole.
Use it when: Prevents marking the angle at the top of the tower.
TOA for towers
Tower Over Away: tan = tower height over distance away.
Use it when: Choosing tan in most single-triangle problems.
Closer means steeper
Walk closer, look higher: the angle grows.
Use it when: Checking figures and assertion-reason items.
Eye, then add
Solve from your eyes, add your eyes.
Use it when: Problems with an observer's height.

Line of Sight and Angle of Elevation: solved examples

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

Example 1: Finding the angle from a ratio · 1 mark · MCQBoard level

A vertical pole of height $63$ m stands on level ground. From a point 18 m from its foot, what is the angle of elevation of its top?
(a) $3 0^{\circ}$
(b) $4 5^{\circ}$
(c) $6 0^{\circ}$
(d) $9 0^{\circ}$

Try it first, then show the solutionSolution

$tan θ = \frac{6 3}{18} = \frac{3}{3} = \frac{1}{3}$ .
So $θ = 3 0^{\circ}$ .

Answer: (a) $3 0^{\circ}$

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

A boy whose eyes are 1.4 m above level ground sees the top of a vertical tree at an angle of elevation of $6 0^{\circ}$ . His horizontal distance from the tree is $53$ m. Find the height of the tree and the length of his line of sight to the top.

Try it first, then show the solutionSolution

Figure: horizontal from eye meets tree at 1.4 m; right angle there. (½)
Part above eye: $53 tan 6 0^{\circ} = 53 \times 3 = 15$ m. (1)
Tree height $= 15 + 1.4 = 16.4$ m. (½)
Line of sight $= \frac{5 3}{c o s 6 0 ^{\circ}} = 103$ m. (1)

Answer: Tree 16.4 m; line of sight $103$ m (about 17.32 m).

Example 3: Angle given with the vertical · 3 marks · Short answerHarder board level

A wire from the top of a vertical pole is pegged to the level ground. The wire makes an angle of $3 0^{\circ}$ with the pole and the peg is 7 m from the foot of the pole. Find the angle of elevation of the top from the peg, the height of the pole and the length of the wire.

Try it first, then show the solutionSolution

Angle with ground $= 9 0^{\circ} - 3 0^{\circ} = 6 0^{\circ}$ , since the pole is vertical. (1)
Height $= 7 tan 6 0^{\circ} = 73$ m. (1)
Wire $= \frac{7}{c o s 6 0 ^{\circ}} = 14$ m. (1)

Answer: Elevation $6 0^{\circ}$ ; height $73$ m (about 12.12 m); wire 14 m.

Example 4: Two elevations from one point (statue on pedestal) · 5 marks · Long answerHarder board level

A statue 1.6 m tall stands on a vertical pedestal. From a point on level ground, the angle of elevation of the top of the statue is $6 0^{\circ}$ and of the top of the pedestal is $4 5^{\circ}$ . Find the height of the pedestal. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Let pedestal height $= h$ , distance from point to foot $= x$ . Draw figure. (1)
From $4 5^{\circ}$ : $tan 4 5^{\circ} = \frac{h}{x}$ , so $x = h$ . (1)
From $6 0^{\circ}$ : $tan 6 0^{\circ} = \frac{h + 1.6}{x}$ , so $3 h = h + 1.6$ . (1)
$h = \frac{1.6}{3 - 1} = \frac{1.6 ( 3 + 1 )}{2} = 0.8 (3 + 1)$ . (1)
$h = 0.8 \times 2.732 = 2.1856$ m. Check: $\frac{3.7856}{2.1856} \approx 1.732$ . (1)

Answer: Pedestal height $0.8 (3 + 1) \approx 2.19$ m.

Example 5: Case study: drone at a sports meet · 4 marks · Case-basedHOTS / competency

At a school sports meet, a drone hovers 30 m above a point O on the level field. A camera operator stands at A on the field with the camera lens 1.5 m above the ground; the horizontal distance AO is $28.5 3$ m. Later the drone rises vertically.
(i) Find the height of the drone above the lens. (1)
(ii) Find the angle of elevation of the drone from the lens. (1)
(iii) The drone rises until the angle of elevation is $6 0^{\circ}$ . Find its new height above the ground and how far it rose. (2)

Try it first, then show the solutionSolution

(i) $30 - 1.5 = 28.5$ m. (1)
(ii) $tan θ = \frac{28.5}{28.5 3} = \frac{1}{3}$ , so $θ = 3 0^{\circ}$ . (1)
(iii) Above lens: $28.5 3 \times 3 = 85.5$ m; height $= 87$ m. (1)
Rise $= 87 - 30 = 57$ m. (1)

Answer: (i) 28.5 m (ii) $3 0^{\circ}$ (iii) 87 m above ground; rose 57 m.

Where students lose marks in Line of Sight and Angle of Elevation

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

Mistake: Marking the angle at the top of the object
Students put the given angle where the line of sight meets the tower.
Wrong working
$tan 6 0^{\circ} = \frac{x}{h}$ with angle at the top, giving $h = \frac{x}{3}$ .
Right working
Angle is at the eye: $tan 6 0^{\circ} = \frac{h}{x}$ , so $h = 3 x$ .
Do this: Elevation always sits at the observer, between horizontal and line of sight.
Usually costs: Most or all marks of the question.
Mistake: Ignoring eye height
The horizontal is drawn from the ground instead of the eye.
Wrong working
Height $= 20 tan 4 5^{\circ} = 20$ m (eyes 1.5 m above ground).
Right working
Height $= 1.5 + 20 tan 4 5^{\circ} = 21.5$ m.
Do this: Draw the horizontal at eye level and add the eye height at the end.
Usually costs: Final answer mark, often 1 mark.
Mistake: Using tan when the slant length is given
Students use tan by habit.
Wrong working
String 50 m at $3 0^{\circ}$ : height $= 50 tan 3 0^{\circ} = \frac{50}{3}$ .
Right working
Height $= 50 sin 3 0^{\circ} = 25$ m.
Do this: Ask which two sides are involved before choosing the ratio.
Usually costs: Method and answer marks.
Mistake: Taking the angle with the vertical as elevation
The question says the wire makes an angle with the pole.
Wrong working
Angle with pole $3 0^{\circ}$ , so $tan 3 0^{\circ} = \frac{h}{d}$ .
Right working
Elevation $= 6 0^{\circ}$ , so $tan 6 0^{\circ} = \frac{h}{d}$ .
Do this: Convert: elevation = 90° minus angle with the vertical.
Usually costs: Whole question.

Exam tips

Always draw and label a figure in 2-, 3- and 5-mark answers; it carries credit.
Write the ratio in words or letters first, e.g. $tan 6 0^{\circ} = \frac{A B}{B C}$ , before substituting.
Leave answers in surd form unless a value of $3$ is given; then rationalise before using decimals.
Check the shape: at $6 0^{\circ}$ height should exceed distance, at $3 0^{\circ}$ it should be less.
State units in the final line, and add eye height when the observer is a person.

Practice questions on Line of Sight and Angle of Elevation

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

Q1 · 2 marks · Very short answerBoard level
A ladder leans against a vertical wall on level ground, making an angle of $6 0^{\circ}$ with the ground. Its top touches the wall 9 m above the ground. Find the length of the ladder and the distance of its foot from the wall.
Hint
Height and ladder: use sin; then tan or cos for the base.
Answer
Ladder $= \frac{9}{s i n 6 0 ^{\circ}} = 63$ m; foot $= \frac{9}{3} = 33$ m.
Q2 · 3 marks · Short answerBoard level
A man whose eyes are 1.8 m above level ground stands 8 m from a house. The angle of elevation of the top of the roof is $4 5^{\circ}$ , and of the tip of an antenna fixed vertically on the roof is $6 0^{\circ}$ . Find the height of the house and the length of the antenna. Use $3 = 1.732$ .
Hint
Work above eye level in both triangles; the antenna is the difference of the two heights.
Answer
Roof above eye level $= 8 tan 4 5^{\circ} = 8$ m, so the house is $8 + 1.8 = 9.8$ m high. Antenna tip above eye level $= 8 tan 6 0^{\circ} = 83 \approx 13.86$ m. Antenna length $= 83 - 8 = 8 (3 - 1) \approx 5.86$ m.
Q3 · 3 marks · Short answerHarder board level
From a point on level ground, the angle of elevation of the top of a vertical tower is $3 0^{\circ}$ . The line of sight to the top is 80 m long. Find the height of the tower and the distance of the point from its foot. How far towards the tower must the person move so that the angle of elevation becomes $6 0^{\circ}$ ? Use $3 = 1.732$ .
Hint
Use sin for height, cos for distance, then tan 60° for the new distance.
Answer
Height $= 80 sin 3 0^{\circ} = 40$ m; distance $= 80 cos 3 0^{\circ} = 403 \approx 69.28$ m. At $6 0^{\circ}$ the distance is $\frac{40}{t a n 6 0 ^{\circ}} = \frac{40}{3} = \frac{40 3}{3} \approx 23.09$ m. Move $403 - \frac{40 3}{3} = \frac{80 3}{3} \approx 46.19$ m.
Q4 · 5 marks · Long answerHarder board level
A vertical tower stands on level ground. The angle of elevation of its top from a point P is $6 0^{\circ}$ . From a point Q, 20 m farther from the tower on the same straight line through P and the foot of the tower, the angle of elevation is $4 5^{\circ}$ . Find the height of the tower. Use $3 = 1.732$ .
Hint
Let PO = x; h = √3x and h = x + 20.
Answer
Let the foot be at distance x m from P and the height be h m. From P: $h = x tan 6 0^{\circ} = 3 x$ . From Q: $h = (x + 20) tan 4 5^{\circ} = x + 20$ . So $3 x = x + 20$ , giving $x = \frac{20}{3 - 1} = 10 (3 + 1)$ . Then $h = x + 20 = 103 + 30 = 10 (3 + 3) \approx 47.32$ m.
Q5 · 3 marks · Short answerHOTS / competency
Riya says: 'Since $6 0^{\circ}$ is twice $3 0^{\circ}$ , the point from which the top of a vertical pole is seen at $6 0^{\circ}$ must be half as far from the pole as the point from which it is seen at $3 0^{\circ}$ .' Is she right? Take the height of the pole as h, find both distances, and compare them. Then find both distances for a 10 m pole.
Hint
Write each distance as $\frac{h}{t a n θ}$ ; the angle doubling does not double $tan θ$ .
Answer
No. At $3 0^{\circ}$ : distance $= \frac{h}{t a n 3 0 ^{\circ}} = h 3$ . At $6 0^{\circ}$ : distance $= \frac{h}{t a n 6 0 ^{\circ}} = \frac{h}{3}$ . Ratio $= \frac{h / 3}{h 3} = \frac{1}{3}$ , so the $6 0^{\circ}$ point is one-third as far, not half. Doubling the angle does not halve the distance, because $tan θ$ is not proportional to $θ$ . For a 10 m pole: $103 \approx 17.32$ m and $\frac{10}{3} \approx 5.77$ m.
Q6 · 1 mark · Assertion-reasonBoard level
Assertion (A): As you walk along level ground straight towards a vertical tower, the angle of elevation of its top increases.
Reason (R): The height of the tower stays fixed while the horizontal distance decreases, so $tan θ = \frac{height}{distance}$ increases.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
What happens to a fraction when its numerator is fixed and its denominator gets smaller?
Answer
(a) Both are true and R explains A: with the height fixed and the distance smaller, $tan θ$ is larger, and for acute angles a larger $tan θ$ means a larger angle.
Q7 · 4 marks · Long answerHarder board level
A person walks at a uniform speed along a straight level road towards a vertical tower 30 m high. The angle of elevation of the top of the tower changes from $3 0^{\circ}$ to $6 0^{\circ}$ in 6 minutes. Ignore the person's height. (i) Find the person's distance from the foot of the tower at the two moments. (ii) Find the walking speed in metres per minute. (iii) How much longer will the person take to reach the foot of the tower? Use $3 = 1.732$ .
Hint
Distance moved = difference of the two horizontal distances; speed is the same throughout.
Answer
(i) At $3 0^{\circ}$ : $\frac{30}{t a n 3 0 ^{\circ}} = 303 \approx 51.96$ m. At $6 0^{\circ}$ : $\frac{30}{3} = 103 \approx 17.32$ m. (1½)
(ii) Distance walked $= 303 - 103 = 203 \approx 34.64$ m in 6 minutes, so speed $= \frac{20 3}{6} = \frac{10 3}{3} \approx 5.77$ m per minute. (1½)
(iii) Remaining distance $103$ m is half of $203$ m, so it takes half of 6 minutes: 3 minutes more. (1)
Q8 · 3 marks · Short answerHarder board level
Two vertical poles of equal height stand on opposite sides of a straight road 80 m wide, one at each edge. From a point on the road between them, the angles of elevation of the tops of the poles are $6 0^{\circ}$ and $3 0^{\circ}$ . Find the height of the poles and the distances of the point from the two poles. Use $3 = 1.732$ .
Hint
Let the point be x m from the pole seen at $6 0^{\circ}$ ; the other pole is then $(80 - x)$ m away. Equate the two expressions for the height.
Answer
Height $h = x tan 6 0^{\circ} = 3 x$ and $h = (80 - x) tan 3 0^{\circ} = \frac{80 - x}{3}$ . So $3 x = 80 - x$ , $x = 20$ . Height $= 203 \approx 34.64$ m; the point is 20 m from one pole and 60 m from the other.
Q9 · 4 marks · Case-basedHOTS / competency
Case study: Meera wants to find the height of her school building without climbing it. Her eyes are 1.5 m above the level ground. From a point A she sees the top of the building at an angle of elevation of $3 0^{\circ}$ . She walks 20 m straight towards the building to a point B, from where the angle of elevation is $6 0^{\circ}$ .
(i) Find the distance of B from the building. (2 marks)
(ii) Find the height of the building. (1 mark)
(iii) Her friend says Meera should simply have stood at the point where the angle of elevation is $4 5^{\circ}$ , since there the height above eye level equals the distance. Where is that point, between A and B or outside them? Justify. (1 mark)
Use $3 = 1.732$ .
Hint
Let B be d m from the building and the top be H m above eye level. Write H twice, once from each point.
Answer
(i) From B: $H = d tan 6 0^{\circ} = 3 d$ . From A: $H = (d + 20) tan 3 0^{\circ} = \frac{d + 20}{3}$ . So $3 d = d + 20$ , $d = 10$ m. (2)
(ii) $H = 103 \approx 17.32$ m above her eyes, so the building is about $17.32 + 1.5 = 18.82$ m high. (1)
(iii) At $4 5^{\circ}$ the distance equals H, about 17.32 m from the building. B is 10 m and A is 30 m away, so this point lies between A and B. (1)

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.What is the line of sight?Show answer
The straight line from the observer's eye to the object.
2.Angle of elevation is measured from?Show answer
The horizontal through the observer's eye, upwards to the line of sight.
3.At $4 5^{\circ}$ elevation, how do height and distance compare?Show answer
They are equal.
4. $tan θ = 3$ gives which angle?Show answer
$6 0^{\circ}$ .
5.What happens to elevation as you walk towards a tower?Show answer
It increases.
6.A boy's eyes are 1.5 m above the ground. The top of a tree is 20 m above his eye level. What is the height of the tree?Show answer
21.5 m.

Line of Sight and Angle of Elevation: frequently asked questions

What is the difference between angle of elevation and angle of depression?

Elevation is when you look up at an object above eye level; depression is when you look down at an object below. Both are measured from the horizontal through your eye. The depression from one point equals the elevation from the other.

Which ratio is used most in elevation problems?

Tan, because most questions link a height with a ground distance. Use sin or cos only when a slanted length such as a string or ladder is involved.

Can the angle of elevation be any value in the board exam?

No. The 2026-27 syllabus uses only 30, 45 and 60 degrees, and at most two right triangles in a problem.

Do I have to draw a figure?

Yes. Examiners expect a neat labelled figure, and it helps you avoid wrong ratios. It often carries part of the marks.

Should I use 1.732 for root 3?

Only if the question gives it or asks for a decimal answer. Otherwise leave the answer in surd form.