Some Applications of Trigonometry uses the ratios from Chapter 8 to find heights and distances that cannot be measured directly: towers, chimneys, kite strings, river widths and the gap between two ships. It defines the line of sight, the angle of elevation and the angle of depression, then solves problems with one right triangle and with two angles of the same object. The skills are drawing the right triangle from words, choosing tan, sin or cos, adding the observer’s height, and handling angles of depression through alternate angles.
Key Concepts
1. Line of Sight, Angle of Elevation and Angle of Depression
A student stands some distance from a minar and looks at its top. The line from the eye to the top of the minar is the line of sight: the line drawn from the eye of an observer to the point in the object viewed by the observer. Its angle with the horizontal through the eye gets one of two names.
- Angle of elevation: the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level. This is the case when we raise our head to look at the object: the top of a tower, a kite, a bird, a balloon.
- Angle of depression: the angle formed by the line of sight with the horizontal when the point being viewed is below the horizontal level. This is the case when we lower our head to look at the point: a girl on a balcony looking at a flower pot on the stair of a temple, a man on a lighthouse looking at a ship.
| Feature | Angle of elevation | Angle of depression |
|---|---|---|
| Position of the object | Above the observer’s eye level | Below the observer’s eye level |
| Head movement | Raise the head | Lower the head |
| Measured from | The horizontal through the eye | The horizontal through the eye |
Both angles are measured from the horizontal, never from a vertical wall or pole.
The link between the two angles. A man at the top P of a tower PB looks down at a point A on the ground at an angle of depression θ. The horizontal PQ through P is parallel to the ground AB, and the line of sight PA is a transversal. So the angle of depression ∠QPA and ∠PAB are alternate angles, and they are equal: the angle of depression of A from P equals the angle of elevation of P from A. NCERT uses this step in Example 6.
How the angle changes with distance. For a fixed height, the nearer the point, the larger the angle. Walking towards a building raises the angle of elevation; a ship sailing away from a lighthouse lowers the angle of depression.
What you need to find a height without climbing. Let CD be the minar, E the student’s feet, A the student’s eye and B the point on the minar level with the eye. NCERT lists three things you need: (i) the distance DE of the student from the foot of the minar, (ii) the angle of elevation ∠BAC of the top, and (iii) the height AE of the student. In the right triangle ABC, BC is opposite to ∠A and AB is adjacent to it, so tan A = BC/AB gives BC. Since BD = AE, the height of the minar is CD = BC + BD = BC + AE.
2. Choosing the Right Trigonometric Ratio
Every problem reduces to one or two right triangles in which you know one side and one angle. Choose the ratio that contains the side you know and the side you want.
| You know / you want | Use | Typical object |
|---|---|---|
| Height and horizontal distance | tan θ = opposite/adjacent (or cot θ) | Tower and ground, pole and shadow, building and ship |
| Height and slant length | sin θ = opposite/hypotenuse | Rope, ladder, slide, kite string |
| Horizontal distance and slant length | cos θ = adjacent/hypotenuse | Foot of a ladder and the ladder |
Values you must know by heart.
| Ratio | 30° | 45° | 60° |
|---|---|---|---|
| sin θ | 1/2 | 1/√2 | √3/2 |
| cos θ | √3/2 | 1/√2 | 1/2 |
| tan θ | 1/√3 | 1 | √3 |
| cot θ | √3 | 1 | 1/√3 |
Useful facts that come from the table.
- At 45°, tan 45° = 1, so the height equals the horizontal distance. Any 45° triangle gives a free equation: height = distance.
- For the same height h, the distance at 30° is h√3 and the distance at 60° is h/√3. The 30° distance is three times the 60° distance, because h√3 ÷ (h/√3) = 3.
- If a pole’s shadow is as long as the pole, tan θ = 1 and the Sun’s altitude is 45°.
Keep surds exact. Write the answer as 15√3 m or 10(√3 − 1) m first. Convert to a decimal only at the last step, and only when the question says “take √3 = 1.73” or “1.732”. Rationalise when a surd is in the denominator: 60/√3 = 60√3/3 = 20√3.
3. Heights and Distances with One Right Triangle
The simplest problems need one right triangle and four steps.
- Sketch: the object vertical, the ground horizontal, the line of sight (or rope, ladder, string) slanting.
- Mark the right angle at the foot of the object and the given angle at the correct vertex.
- Pick the ratio that links the known side and the unknown side, and solve.
- Add or subtract extra lengths, such as the observer’s height.
NCERT Example 1 (tower). A tower stands vertically on the ground. From a point on the ground 15 m from the foot of the tower, the angle of elevation of the top is 60°. Let AB be the tower and C the point, with the triangle right-angled at B. tan 60° = AB/BC, so √3 = AB/15 and AB = 15√3 m.
NCERT Example 2 (electrician and ladder). A pole AD is 5 m high. The electrician must reach a point B that is 1.3 m below the top, so BD = 5 − 1.3 = 3.7 m. The ladder BC is inclined at 60° to the horizontal. The ladder is the hypotenuse and BD is opposite the 60° angle, so use sin: BD/BC = sin 60° = √3/2. So BC = 3.7 × 2/√3 = 7.4/1.73 ≈ 4.28 m. For the foot of the ladder, DC/BD = cot 60° = 1/√3, so DC = 3.7/1.73 ≈ 2.14 m. The ladder should be about 4.28 m long, with its foot about 2.14 m from the pole.
NCERT Example 3 (observer’s height). An observer 1.5 m tall stands 28.5 m from a chimney, and the angle of elevation of the top from her eyes is 45°. Let AB be the chimney, D the observer’s eye and E the point on AB level with the eye. DE = 28.5 m and tan 45° = AE/DE, so AE = 28.5 m. Height of chimney = AE + BE = 28.5 + 1.5 = 30 m.
The observer’s height rule. When the angle is measured “from the eyes” of a person of height a, find the part above eye level with the triangle, then add a. When the object’s height is given and the angle is from the eyes, subtract a first. In Exercise 9.1 Q6, the building is 30 m and the boy is 1.5 m, so the triangle uses 30 − 1.5 = 28.5 m.
Broken tree. The standing part is the vertical side, the broken part is the hypotenuse, and the ground distance is the base. The original height is standing part + broken part.
4. Two Angles of Elevation of the Same Object
Many questions give two angles of the same object. This happens in four common ways:
- An observer walks towards or away from a building, and the angle of elevation changes.
- A tower’s shadow is longer at one altitude of the Sun than at another.
- One point sees two things on the same vertical line: a building and a flagstaff on it, a pedestal and a statue on it.
- A point lies between two objects, such as two poles on opposite sides of a road.
Method. Call the height h and a distance x. Write one tan equation per right triangle, then substitute one into the other.
NCERT Example 4 (building and flagstaff). From a point P, the angle of elevation of the top B of a 10 m building AB is 30°, and the angle of elevation of the top D of a flagstaff on the building is 45°. In right △PAB, tan 30° = AB/AP, so 1/√3 = 10/AP and AP = 10√3 m = 17.32 m. Let the flagstaff BD = x m, so AD = (10 + x) m. In right △PAD, tan 45° = AD/AP, so 1 = (10 + x)/(10√3). Then x = 10√3 − 10 = 10(√3 − 1) = 7.32 m.
The Sun’s altitude. The altitude of the Sun is the angle of elevation of the Sun. For a tower of height h, the angle of elevation of its top from the tip of its shadow equals the Sun’s altitude, so the shadow length is h cot θ. The lower the Sun, the longer the shadow.
NCERT Example 5 (two shadows). The shadow of a tower is 40 m longer when the Sun’s altitude is 30° than when it is 60°. Let the tower AB = h m and the shorter shadow BC = x m, so the longer shadow DB = (x + 40) m. In △ABC, tan 60° = h/x, so h = √3 x …(1). In △ABD, tan 30° = h/(x + 40), so 1/√3 = h/(x + 40) …(2). Putting (1) into (2): √3 × √3 x = x + 40, so 3x = x + 40 and x = 20. Then h = 20√3 m.
Shortcut for 30° and 60°. For the same height h, the gap between the two points is h√3 − h/√3 = 2h/√3. In Example 5 the gap is 40, so 2h/√3 = 40 and h = 20√3, the same answer in one line. Use it to check; show the full working in the exam.
5. Problems Using Angles of Depression
Angles of depression are measured at the top, between the horizontal and a line of sight going down. The first move every time is to transfer that angle to the ground point, where it becomes an angle of elevation (alternate angles). After that the problem is solved exactly like an elevation problem.
NCERT Example 6 (two buildings). From the top P of a multi-storeyed building PC, the angles of depression of the top B and the bottom A of an 8 m building AB are 30° and 45°. Draw BD horizontal from B to meet PC at D. Then ∠PBD = 30° and ∠PAC = 45° (alternate angles). In right △PBD, PD/BD = tan 30° = 1/√3, so BD = PD√3. In right △PAC, PC/AC = tan 45° = 1, so PC = AC. Also PC = PD + DC with DC = AB = 8 m, and AC = BD. So PD + 8 = PD√3, giving PD = 8/(√3 − 1) = 8(√3 + 1)/2 = 4(√3 + 1) m. The height PC = 4(√3 + 1) + 8 = 4(3 + √3) m, and the distance between the buildings is also 4(3 + √3) m.
NCERT Example 7 (river and bridge). From a point P on a bridge 3 m above the banks, the angles of depression of the banks on opposite sides are 30° and 45°. Let A and B be the banks and D the point on AB directly below P, so PD = 3 m. In right △APD, ∠A = 30° and tan 30° = PD/AD, so AD = 3√3 m. In right △PBD, ∠B = 45°, so BD = PD = 3 m. Width AB = AD + DB = 3√3 + 3 = 3(1 + √3) m.
Same side or opposite sides.
| Arrangement | What to do with the two distances | Example |
|---|---|---|
| Both points on the same side of the tower (one behind the other) | Subtract: gap = far distance − near distance | Two ships seen from a lighthouse (Exercise 9.1 Q13) |
| Points on opposite sides of the tower or bridge | Add: total = sum of the two distances | Two banks of a river (NCERT Example 7), point between two poles (Exercise 9.1 Q10) |
Elevation and depression from one point. From the top of a short building you may see the top of a taller tower (elevation) and its foot (depression). Draw the horizontal from your eye to the tower. It splits the tower into a lower part equal to the building’s height and an upper part found from the elevation angle. Exercise 9.1 Q12 uses this.
6. A Complete Method, Including Moving Objects
- Sketch and mark every given length and angle.
- Find the right triangles. Name them with letters. Mark the right angle.
- Move depression angles to the ground point as equal elevation angles.
- Adjust for eye height if the angle is measured from the eyes.
- Write one equation per triangle using the ratio that contains the known and unknown sides.
- Solve, keeping surds exact. Rationalise the denominator.
- State the answer with units, and in decimals if the question gives a value for √3.
Moving objects. Some problems add speed and time. A car approaching a tower at a uniform speed changes its angle of depression. A balloon moving horizontally changes its angle of elevation. Find each position with a tan equation, subtract to get the distance travelled, then use speed = distance/time.
For the car in Exercise 9.1 Q15, with tower height h, the distance at 30° is h√3 and at 60° is h/√3. The car covers h√3 − h/√3 = 2h/√3 in 6 seconds and still has h/√3 to go. The remaining distance is half the distance already covered, so at the same speed it takes half the time: 3 seconds. The height h cancels out, so the answer does not depend on how tall the tower is.
Sense check. At 60° the height exceeds the horizontal distance; at 30° it is less. A slant length is always longer than the height it reaches.
Formula and Theorem Sheet
| Situation | Relation | Remember |
|---|---|---|
| Height h, horizontal distance d, elevation θ | tan θ = h/d, so h = d tan θ and d = h cot θ | The most used relation in the chapter |
| Slant length L making θ with the ground | h = L sin θ, d = L cos θ | Rope, ladder, slide, kite string |
| Observer of height a, angle from the eyes | Object height = d tan θ + a | Subtract a first if the object height is given |
| Angle of depression θ from height h | Ground distance = h cot θ | Equal to the angle of elevation at the ground point (alternate angles) |
| Same height, angles 30° and 60° | Distances h√3 and h/√3; gap = 2h/√3 | The 30° distance is 3 times the 60° distance |
| Same height, angles 30° and 45° | Distances h√3 and h; gap = h(√3 − 1) | Two ships, one behind the other |
| Object on a base (flagstaff, statue, tower on a building) | Base height = d tan α, total = d tan β, object = difference | Same point, same horizontal distance d |
| Shadow of height h, Sun’s altitude θ | Shadow = h cot θ | Lower Sun, longer shadow |
| Rationalising | 1/(√3 − 1) = (√3 + 1)/2; 1/(√3 + 1) = (√3 − 1)/2 | NCERT Example 6 and Exercise 9.1 Q8 |
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Important Definitions
- Line of sight: the line drawn from the eye of an observer to the point in the object viewed by the observer.
- Angle of elevation: the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level, that is, when we raise our head to look at the object.
- Angle of depression: the angle formed by the line of sight with the horizontal when the point being viewed is below the horizontal level, that is, when we lower our head to look at the point being viewed.
- Horizontal level: the level line through the observer’s eye, parallel to the level ground. Both angles are measured from it.
- Altitude of the Sun: the angle of elevation of the Sun. It equals the angle of elevation of the top of a vertical object from the tip of its shadow.
- Heights and distances: the use of trigonometric ratios to find the height or length of an object, or the distance between two distant objects, without measuring them directly.
Solved Examples (NCERT-Based)
Example 1: The broken tree (NCERT Exercise 9.1, Q2)
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Solution: Let B be the foot of the tree, A the point where it breaks, and C where the top touches the ground. AB is the standing part, AC the broken part, BC = 8 m and ∠ACB = 30°.
tan 30° = AB/BC ⇒ 1/√3 = AB/8 ⇒ AB = 8/√3 m.
cos 30° = BC/AC ⇒ √3/2 = 8/AC ⇒ AC = 16/√3 m.
Height of tree = AB + AC = 8/√3 + 16/√3 = 24/√3 = 24√3/3 = 8√3 m.
The tree was 8√3 m ≈ 13.86 m high.
Example 2: Two slides in a park (NCERT Exercise 9.1, Q3)
A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Solution: The slide is the hypotenuse and its height is opposite the angle, so use sin.
Small slide: sin 30° = 1.5/L1 ⇒ 1/2 = 1.5/L1 ⇒ L1 = 3 m.
Steep slide: sin 60° = 3/L2 ⇒ √3/2 = 3/L2 ⇒ L2 = 6/√3 = 2√3 m.
The slides should be 3 m and 2√3 m (≈ 3.46 m) long.
Example 3: The boy walking towards a building (NCERT Exercise 9.1, Q6)
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution: The angles are measured from his eyes, so the height above eye level is 30 − 1.5 = 28.5 m. Let the horizontal distances from the building be d1 at 30° and d2 at 60°.
tan 30° = 28.5/d1 ⇒ d1 = 28.5√3 m.
tan 60° = 28.5/d2 ⇒ d2 = 28.5/√3 m.
Distance walked = d1 − d2 = 28.5(√3 − 1/√3) = 28.5 × 2/√3 = 57/√3 = 19√3 m.
He walked 19√3 m ≈ 32.91 m.
Example 4: Transmission tower on a building (NCERT Exercise 9.1, Q7)
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Solution: Let P be the point, B the foot of the building, C the top of the building (bottom of the tower) and D the top of the tower. BC = 20 m; let CD = h m.
In △PBC: tan 45° = BC/PB ⇒ 1 = 20/PB ⇒ PB = 20 m.
In △PBD: tan 60° = BD/PB ⇒ √3 = (20 + h)/20 ⇒ 20 + h = 20√3 ⇒ h = 20(√3 − 1).
The tower is 20(√3 − 1) m ≈ 14.64 m high.
Example 5: Statue on a pedestal (NCERT Exercise 9.1, Q8)
A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Solution: Let the pedestal be h m high and the point be d m from its foot.
Top of pedestal: tan 45° = h/d ⇒ d = h.
Top of statue: tan 60° = (h + 1.6)/d ⇒ √3 h = h + 1.6 ⇒ h(√3 − 1) = 1.6.
h = 1.6/(√3 − 1) = 1.6(√3 + 1)/(3 − 1) = 0.8(√3 + 1).
The pedestal is 0.8(√3 + 1) m ≈ 2.19 m high.
Example 6: Building and tower (NCERT Exercise 9.1, Q9)
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution: Let the distance between the feet of the building and the tower be d m and the building be h m high.
From the foot of the building: tan 60° = 50/d ⇒ d = 50/√3 m.
From the foot of the tower: tan 30° = h/d ⇒ h = d/√3 = (50/√3) × (1/√3) = 50/3.
The building is 50/3 m = 16 2/3 m ≈ 16.67 m high.
Example 7: Two poles across a road (NCERT Exercise 9.1, Q10)
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
Solution: Let each pole be h m high and the point be x m from the pole seen at 60°. Then it is (80 − x) m from the other pole.
tan 60° = h/x ⇒ h = √3 x …(1)
tan 30° = h/(80 − x) ⇒ h = (80 − x)/√3 …(2)
From (1) and (2): √3 x = (80 − x)/√3 ⇒ 3x = 80 − x ⇒ x = 20. Then h = 20√3.
Each pole is 20√3 m ≈ 34.64 m high. The point is 20 m from one pole and 60 m from the other.
Example 8: TV tower across a canal (NCERT Exercise 9.1, Q11)
A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.
Solution: Let the tower be h m high and the canal x m wide. The second point is 20 m further away, at (x + 20) m.
tan 60° = h/x ⇒ h = √3 x …(1)
tan 30° = h/(x + 20) ⇒ h = (x + 20)/√3 …(2)
From (1) and (2): 3x = x + 20 ⇒ x = 10. Then h = 10√3.
The tower is 10√3 m ≈ 17.32 m high and the canal is 10 m wide.
Example 9: Cable tower from a building (NCERT Exercise 9.1, Q12)
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution: Let A be the top of the building, B its foot, C the foot of the tower and D its top. Draw AE horizontal from A to meet CD at E. Then EC = AB = 7 m and AE = BC.
Depression of C is 45°, so ∠ACB = 45° (alternate angles). tan 45° = AB/BC ⇒ BC = 7 m, so AE = 7 m.
In △AED: tan 60° = DE/AE ⇒ DE = 7√3 m.
Height of tower = DE + EC = 7√3 + 7 = 7(√3 + 1).
The tower is 7(√3 + 1) m ≈ 19.12 m high.
Example 10: Two ships from a lighthouse (NCERT Exercise 9.1, Q13)
As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Solution: Each angle of depression equals the angle of elevation of the top of the lighthouse from that ship. The ship at 45° is nearer.
Near ship: tan 45° = 75/d1 ⇒ d1 = 75 m.
Far ship: tan 30° = 75/d2 ⇒ d2 = 75√3 m.
Both ships are on the same side, so subtract: d2 − d1 = 75√3 − 75 = 75(√3 − 1).
The ships are 75(√3 − 1) m ≈ 54.9 m apart.
Example 11: The balloon (NCERT Exercise 9.1, Q14)
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.
Solution: The angles are from her eyes, so the balloon is 88.2 − 1.2 = 87 m above eye level in both positions.
First position: tan 60° = 87/d1 ⇒ d1 = 87/√3 = 29√3 m.
Second position: tan 30° = 87/d2 ⇒ d2 = 87√3 m.
Distance travelled = d2 − d1 = 87√3 − 29√3 = 58√3.
The balloon travelled 58√3 m ≈ 100.46 m.
Example 12: The car on the highway (NCERT Exercise 9.1, Q15)
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Solution: Let the tower be h m high. The depression angles become elevation angles at the car.
At 30°: distance = h/tan 30° = h√3. At 60°: distance = h/tan 60° = h/√3.
Distance covered in 6 s = h√3 − h/√3 = (3h − h)/√3 = 2h/√3.
Speed = (2h/√3) ÷ 6 = h/(3√3) m/s.
Time for the remaining h/√3 = (h/√3) ÷ (h/(3√3)) = 3 s.
The car takes 3 seconds more to reach the foot of the tower.
Competency-Based Questions (with answers)
1. Case-based: The lighthouse keeper
A lighthouse stands on a rock at the edge of the sea. Its lamp is 60 m above sea level. The keeper watches a boat sailing straight towards the lighthouse. At first the angle of depression of the boat from the lamp is 30°. Later it is 60°.
(a) Name the angle equal to the angle of depression at the boat. The angle of elevation of the lamp from the boat, because the two are alternate angles between the horizontal through the lamp and the sea surface.
(b) Find the first distance of the boat from the foot of the lighthouse. d1 = 60/tan 30° = 60√3 m.
(c) How far did the boat sail between the two sightings? d2 = 60/tan 60° = 60/√3 = 20√3 m. Distance = 60√3 − 20√3 = 40√3 m ≈ 69.28 m.
2. Case-based: A ramp to a stage
A school builds a straight ramp from the ground to a stage that is 0.9 m above the ground. The ramp makes an angle of 30° with the ground.
(a) Which ratio links the ramp length with the height? sin 30°, since the ramp is the hypotenuse and the height is opposite the angle.
(b) Find the length of the ramp. sin 30° = 0.9/L ⇒ L = 1.8 m.
(c) How far from the stage does the ramp start? d = L cos 30° = 1.8 × √3/2 = 0.9√3 ≈ 1.56 m (taking √3 = 1.73).
3. Source-based: What you need to measure a minar
NCERT says that to find the height CD of a minar you need the distance DE of the student from its foot, the angle of elevation ∠BAC of the top, and the height AE of the student. BD = AE, and tan A or cot A gives BC.
(a) Why is tan A chosen and not sin A? In △ABC the known side AB (equal to DE) and the wanted side BC are the two sides at the right angle. tan A = BC/AB contains both. sin A would need the hypotenuse AC, which is unknown.
(b) If DE = 40 m, ∠BAC = 45° and AE = 1.6 m, find the height of the minar. BC = 40 tan 45° = 40 m. CD = BC + BD = 40 + 1.6 = 41.6 m.
4. Assertion-Reason
Assertion (A): The angle of depression of a boat from the top of a cliff is equal to the angle of elevation of the top of the cliff from the boat.
Reason (R): The horizontal through the observer’s eye is parallel to the level ground, so the two angles are alternate angles.
Choose: (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.
Answer: (a). Both statements are true, and the parallel-lines argument in R is exactly why the angles are equal.
5. Assertion-Reason
Assertion (A): If the length of the shadow of a vertical pole is equal to its height, the Sun’s altitude is 45°.
Reason (R): sin 45° = cos 45°.
Choose: (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.
Answer: (b). A is true because tan θ = height/shadow = 1 gives θ = 45°. R is also true, since both equal 1/√2. R does not explain A; the explanation uses tan θ = 1.
6. Assertion-Reason
Assertion (A): As an observer walks away from a tower, the angle of elevation of its top increases.
Reason (R): For a fixed height h, the distance d = h cot θ, and cot θ decreases as θ increases.
Choose: (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.
Answer: (d). A is false: walking away increases d, so cot θ must increase, which means θ decreases. R is true.
7. Error analysis: The missing eye height
A student solves: “A 1.6 m tall man stands 20 m from a tree. The angle of elevation of the top of the tree from his eyes is 45°. Height of the tree = 20 tan 45° = 20 m.” Find the error and correct it.
Answer: The angle is measured from his eyes, so the triangle gives only the part of the tree above eye level, 20 m. The man’s height must be added. Height of the tree = 20 + 1.6 = 21.6 m.
8. Competency MCQ: The ladder
A ladder 10 m long leans against a vertical wall and makes an angle of 60° with the ground. How high up the wall does it reach?
(a) 5 m (b) 5√3 m (c) 10√3 m (d) 10/√3 m
Answer: (b). Height = 10 sin 60° = 10 × √3/2 = 5√3 m. Option (a) is the distance of the foot from the wall, 10 cos 60° = 5 m. The height (≈ 8.66 m) must be less than the ladder, which rules out (c).
Important Questions for Board Exams
1-Mark Questions
- Define the angle of depression. It is the angle formed by the line of sight with the horizontal when the point being viewed is below the horizontal level, as when we lower our head to look at it.
- The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower. (NCERT Exercise 9.1, Q4) h = 30 tan 30° = 30/√3 = 10√3 m ≈ 17.32 m.
- A pole 6 m high casts a shadow 2√3 m long. Find the Sun’s altitude. tan θ = 6/(2√3) = √3, so θ = 60°.
- If the height of a tower and the distance of a point from its foot are equal, find the angle of elevation of the top from that point. tan θ = 1, so θ = 45°.
2-Mark Questions
- A ladder makes an angle of 60° with the ground and its foot is 2.5 m from the wall. Find the length of the ladder. cos 60° = 2.5/L ⇒ 1/2 = 2.5/L ⇒ L = 5 m.
- From the top of a 50 m high cliff, the angle of depression of a boat is 30°. Find the distance of the boat from the foot of the cliff. The angle of elevation at the boat is 30°. d = 50/tan 30° = 50√3 m ≈ 86.6 m.
- The angles of elevation of the top of a tower from two points 4 m and 9 m from its base, on the same side and in the same straight line, are complementary. Prove that the tower is 6 m high. Let the angles be θ and 90° − θ. tan θ = h/4 and tan(90° − θ) = cot θ = h/9. Multiplying, tan θ × cot θ = h2/36, so 1 = h2/36 and h = 6 m.
3-Mark Questions
- A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string. (NCERT Exercise 9.1, Q5) sin 60° = 60/L ⇒ √3/2 = 60/L ⇒ L = 120/√3 = 40√3 m ≈ 69.28 m.
- The angle of elevation of the top of a tower from a point on the ground is 30°. After walking 20 m towards the tower, the angle becomes 60°. Find the height of the tower. Distances are h√3 and h/√3. h√3 − h/√3 = 20 ⇒ 2h/√3 = 20 ⇒ h = 10√3 m ≈ 17.32 m.
- From the top of a 60 m high building, the angles of depression of the top and the bottom of a tower are 30° and 60°. Find the height of the tower. From the bottom (60°): distance d = 60/tan 60° = 20√3 m. From the top (30°): the building rises above the tower’s top by d tan 30° = 20√3 × 1/√3 = 20 m. Tower = 60 − 20 = 40 m.
5-Mark Questions
- The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (NCERT Example 6) Let the multi-storeyed building be PC and the small building AB, and draw BD horizontal to meet PC at D. From 45°: PC = AC. From 30°: BD = PD√3. Since AC = BD and PC = PD + 8, PD + 8 = PD√3, so PD = 8/(√3 − 1) = 4(√3 + 1). Height PC = 4(√3 + 1) + 8 = 4(3 + √3) m ≈ 18.93 m. Distance AC = PC = 4(3 + √3) m.
- A man on the deck of a ship is 10 m above the water level. He sees the top of a cliff at an angle of elevation of 45° and its base at an angle of depression of 30°. Find the distance of the cliff from the ship and the height of the cliff. Base (30°): d = 10/tan 30° = 10√3 m ≈ 17.32 m. Top (45°): the part of the cliff above his eye level = d tan 45° = 10√3 m. Height of cliff = 10 + 10√3 = 10(√3 + 1) m ≈ 27.32 m.
Common Mistakes and Examiner Tips
- Measuring the angle from the vertical. Angles of elevation and depression are always measured from the horizontal. If a question gives the angle a rope makes with a pole, the angle with the ground is 90° minus that angle.
- Placing the angle of depression at the wrong vertex. The angle of depression sits at the top, between the horizontal and the line of sight. Inside the triangle, use the equal angle of elevation at the ground point.
- Forgetting the observer’s height. “From her eyes” means the triangle starts at eye level. Add the eye height to the answer, or subtract it from a given height first (Q6 and Q14 of Exercise 9.1).
- Using sin when tan is needed. Height and ground distance are the two legs, so use tan. Use sin only when the hypotenuse (rope, ladder, string, slide) is involved.
- Giving only the standing part of a broken tree. The tree’s height is the standing part plus the broken part. In Exercise 9.1 Q2 it is 8/√3 + 16/√3 = 8√3 m.
- Adding when you should subtract. Two ships on the same side of a lighthouse: subtract the distances. Two banks of a river, or a point between two poles: add them.
- Swapping the two angles. The larger angle belongs to the nearer point. At 60° the observer is closer than at 30°.
- Using decimals too early. Writing 1/√3 as 0.577 in the first line causes rounding errors. Keep surds to the end and convert once, with the value the question gives.
- Leaving a surd in the denominator. Write 40√3, not 120/√3. For 8/(√3 − 1), multiply by (√3 + 1)/(√3 + 1) to get 4(√3 + 1).
- Skipping the figure. A clear labelled sketch shows your method, even if the arithmetic slips. Name every point you use in an equation.
- Answering the wrong quantity. Exercise 9.1 Q7 asks for the tower on the building, 20(√3 − 1) m, not the total height 20√3 m. Q15 asks for the time after the second sighting, 3 s, not the speed. Reread the last line of the question, and give units.
Quick Revision Points
- The line of sight is the line from the observer’s eye to the point being viewed.
- Angle of elevation: object above eye level, head raised.
- Angle of depression: object below eye level, head lowered.
- Both angles are measured from the horizontal through the eye.
- Angle of depression from the top = angle of elevation from the ground point (alternate angles).
- Height and horizontal distance: use tan. Height and slant length: use sin. Distance and slant length: use cos.
- tan 30° = 1/√3, tan 45° = 1, tan 60° = √3.
- sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2.
- At 45°, height = horizontal distance.
- h = d tan θ and d = h cot θ.
- Angle measured from the eyes: add the observer’s height at the end.
- For the same height, the 30° distance is three times the 60° distance.
- Two angles: write one equation per triangle and solve them together.
- Same side: subtract distances. Opposite sides: add them.
- Broken tree height = standing part + broken part.
- Shadow length = h cot (Sun’s altitude). Lower Sun, longer shadow.
- Object on a base: find the base and the total from the same point, then subtract.
- Moving objects: distance travelled = difference of the two distances, then speed = distance/time.
- Keep surds exact and rationalise: 1/(√3 − 1) = (√3 + 1)/2.
- Answer checks: at 60° the height exceeds the distance; a slant length always exceeds the height.
Weightage in Board Exams
Some Applications of Trigonometry builds directly on Introduction to Trigonometry (Chapter 8) and uses its ratios and standard values throughout. Check the current CBSE course structure for the marks given to this part of the syllabus. This chapter supplies the application questions on heights and distances.
| Question type | What is usually asked |
|---|---|
| MCQ and Assertion-Reason | Definitions of elevation and depression; the Sun’s altitude from a pole and its shadow; one-step height or distance with a standard angle |
| Short answers | One right triangle: tower, ladder, kite string, slide, cliff and boat |
| Long answers | Two angles: moving observer, two ships, flagstaff or statue on a base, two buildings, a moving car |
| Case-based | A real setting such as a lighthouse, bridge or ramp, split into parts that build from one triangle to two |
A good answer shows a labelled figure, the ratio chosen, one equation per triangle and the final answer with units. Practise every NCERT example and all fifteen questions of Exercise 9.1.
Class 10 Maths Β· Chapter 9 β swipe through all 10 cards to understand the whole chapter.
Line of Sight
The line drawn from the eye of an observer to the point in the object viewed.
Every angle in this chapter is measured between this line and the horizontal.
- Draw a horizontal line through the observer’s eye first.
- The line of sight is the slanting side of your right triangle.
- The object is always drawn as a vertical line.
Angle of Elevation
The angle between the line of sight and the horizontal when the object is above eye level.
Nearer point, larger angle of elevation.
- We raise our head to see the object.
- Tower 15 m away at 60Β°: height = 15β3 m.
- Walking towards a building increases the angle.
Angle of Depression
The angle between the line of sight and the horizontal when the object is below eye level.
Equal because they are alternate angles between parallel horizontals.
- We lower our head to see the object.
- Mark the equal angle at the ground point, then solve.
- Distance of object = h cot ΞΈ.
Choosing tan, sin or cos
Use the ratio that contains the side you know and the side you want.
sin 30Β° = 1/2, sin 60Β° = β3/2, cos 60Β° = 1/2.
- Height and ground distance: tan.
- Height and rope, ladder or string: sin.
- Ground distance and slant length: cos.
Observer’s Height
An angle measured from the eyes gives only the part above eye level.
If the object’s height is given, subtract the eye height first.
- 1.5 m observer, 28.5 m away, 45Β°: chimney = 30 m.
- 30 m building, 1.5 m boy: work with 28.5 m.
- 88.2 m balloon, 1.2 m girl: work with 87 m.
Ladders, Kites and Trees
Many problems need a single right triangle and one ratio.
A broken tree’s height is standing part + broken part.
- 20 m rope at 30Β°: pole = 10 m.
- Kite at 60 m, string at 60Β°: string = 40β3 m.
- Broken tree, 8 m base, 30Β°: height = 8β3 m.
Two Angles, One Object
Write one tan equation per triangle and solve them together.
Gap between the 30Β° and 60Β° points = 2h/β3.
- Shadow 40 m longer at 30Β° than at 60Β°: h = 20β3 m.
- Flagstaff on 10 m building, 30Β° and 45Β°: flagstaff = 7.32 m.
- Tower on 20 m building, 45Β° and 60Β°: tower = 20(β3 β 1) m.
Same Side or Opposite Sides
The arrangement of the two points decides how to combine the distances.
Draw the figure to see which case you have.
- Two ships from a 75 m lighthouse at 30Β° and 45Β°: 75(β3 β 1) m.
- River from a 3 m bridge at 30Β° and 45Β°: 3(1 + β3) m.
- Point between two poles on an 80 m road: 20 m and 60 m.
Speed and Time
Find both positions with tan, subtract, then use speed = distance / time.
In the highway problem the tower’s height cancels out.
- Car seen at 30Β° then 60Β° after 6 s: 3 s more to the foot.
- Balloon at 87 m above eye, 60Β° then 30Β°: moves 58β3 m.
- Keep h as a letter; it often cancels.
Surds and Answer Checks
Keep surds exact, rationalise, and convert to decimals only at the end.
Use only the value of β3 the question gives (1.73 or 1.732).
- At 60Β° the height is more than the ground distance.
- A slant length is always longer than the height it reaches.
- State the answer with units in a full sentence.
π Practice Some Applications of Trigonometry - 10 board questions
CBSE previous-year and competency-based Β· with answers & explanations
Start βClose β
Why D: Step 1 (tan links height and ground distance): tan ΞΈ = 30 / (10β3) = 3/β3 = β3. Step 2 (standard value tan 60Β° = β3): ΞΈ = 60Β°.
Why not A: 30Β° comes from distance / height = 10β3/30 = 1/β3, the ratio turned upside down.
Why not B: 45Β° needs the distance to equal the height, but 10β3 β 17.3 m is less than 30 m.
Why not C: 90Β° would mean the car is at the base of the tower, which it is not.
Remember: tan ΞΈ = height / distance, height on top.
Why B: B is the foot of the pole, so AB is the ground distance and the wire AC is the hypotenuse. Step 1 (cos links adjacent side and hypotenuse): cos 60Β° = AB / AC. Step 2 (standard value cos 60Β° = 1/2): AC = 5β3 / (1/2) = 10β3 m.
Why not A: 10 m is 5β3 / cos 30Β°, using the wrong angle.
Why not C: 15 m is 5β3 Γ tan 60Β° = 5β3 Γ β3, the height of the pole BC, not the wire.
Why not D: (5/2)β3 comes from multiplying by cos 60Β° instead of dividing, which makes the hypotenuse shorter than a leg.
Remember: The hypotenuse is always the longest side; if your answer is shorter than a leg, you divided the wrong way.
Why C: A is true: the ladder is the hypotenuse, β(6Β² + 8Β²) = β100 = 10 m. R is false: tan ΞΈ = 8/6 = 4/3 β 1.33, while tan 60Β° = β3 β 1.73, so the angle is about 53Β°, not 60Β°.
Why not A: R is false, so it cannot explain A; the length also comes from Pythagoras, not from an angle.
Why not B: B needs R to be true, but 8/6 is not β3.
Why not D: A is true, since 6Β² + 8Β² = 10Β².
Remember: A 6-8-10 ladder does not sit at 60Β°; check tan before trusting a standard angle.
Why B: Step 1 (sin links opposite side and hypotenuse): sin 30Β° = 150 / L. Step 2 (standard value sin 30Β° = 1/2): L = 150 Γ 2 = 300 m.
Why not A: 100β3 = 300/β3 comes from dividing 150 by cos 30Β° = β3/2 instead of by sin 30Β°.
Why not C: 150β2 m would be the string length at 45Β°, where sin 45Β° = 1/β2.
Why not D: 150β3 m is 150 / tan 30Β°, the horizontal ground distance, not the string.
Remember: Height and string: use sin; height and ground: use tan.
Why B: Step 1 (tan links height and shadow): tan ΞΈ = height / shadow = h / h = 1. Step 2 (standard value): tan 45Β° = 1, so ΞΈ = 45Β°.
Why not A: At 30Β°, tan ΞΈ = 1/β3, so the shadow would be β3 times the height, longer than the tower.
Why not C: At 60Β°, tan ΞΈ = β3, so the shadow would be shorter than the tower.
Why not D: At 90Β° the Sun is overhead and there is no shadow at all.
Remember: Shadow equals height: 45Β°.
Why B: Step 1 (tan links height and ground distance): tan 60Β° = h / 30. Step 2 (standard value tan 60Β° = β3): h = 30β3 m, about 51.96 m.
Why not A: 10β3 = 30/β3 comes from using tan 30Β° instead of tan 60Β°.
Why not C: 60 m is 30 / cos 60Β°, the length of the line of sight, not the height.
Why not D: 30 m would need tan ΞΈ = 1, which is true only at 45Β°.
Remember: Height = distance Γ tan(angle).
Why A: Step 1 (tan links height and shadow): tan ΞΈ = 6 / (2β3) = 3/β3 = β3. Step 2 (standard value tan 60Β° = β3): ΞΈ = 60Β°.
Why not B: 45Β° needs tan ΞΈ = 1, which means shadow equal to height, but 2β3 β 3.46 m is shorter than 6 m.
Why not C: 30Β° comes from shadow / height = 2β3/6 = 1/β3, the ratio turned upside down.
Why not D: At 90Β° the Sun is overhead and there is no shadow at all.
Remember: Shadow shorter than the pole means the Sun is above 45Β°.
Why A: The horizontal through the observer on the cliff is parallel to the sea level, and the line of sight joining the cliff top to the boat is a transversal. The angle of depression at the top and the angle of elevation at the boat are alternate angles, so both equal 35Β°.
Why not B: 55Β° is 90Β° – 35Β°, the angle with the vertical cliff face, not with the horizontal.
Why not C: 145Β° is 180Β° – 35Β°, a co-interior angle, which is not the angle of elevation.
Why not D: 70Β° doubles the angle; there is no reason to add the two equal angles together.
Remember: Depression from the top equals elevation from the bottom.
Why D: Step 1 (tan links height and shadow): tan ΞΈ = height / shadow = h / h = 1. Step 2 (standard value): tan 45Β° = 1, so ΞΈ = 45Β°.
Why not A: At 30Β°, tan ΞΈ = 1/β3, so the shadow would be β3 times the height, longer than the pole.
Why not B: At 60Β°, tan ΞΈ = β3, so the shadow would be shorter than the pole.
Why not C: At 90Β° the Sun is overhead and there is no shadow at all.
Remember: Shadow equals height: 45Β°.
Why A: Step 1 (tan for the standing part): standing part = 8 tan 30Β° = 8/β3 m. Step 2 (cos for the broken part, the hypotenuse): cos 30Β° = 8 / broken part, so broken part = 8 Γ 2/β3 = 16/β3 m. Step 3 (full height = sum): 8/β3 + 16/β3 = 24/β3 = 8β3 m, about 13.86 m.
Why not B: 8/β3 m is only the standing stump; the broken part must be added.
Why not C: 16/β3 m is only the broken part; the standing stump must be added.
Why not D: 16 m = 8 / sin 30Β° uses sin for a ratio of ground distance to hypotenuse, which needs cos.
Remember: Broken tree height = stump + broken part.
Chapter Navigation
Previous: Introduction to Trigonometry Class 10 Notes
Next: Circles Class 10 Notes
Related Chapters in Class 10 Maths
- Real Numbers Class 10 Notes
- Polynomials Class 10 Notes
- Pair of Linear Equations in Two Variables Class 10 Notes
- Quadratic Equations Class 10 Notes
- Arithmetic Progressions Class 10 Notes
- Triangles Class 10 Notes
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Frequently Asked Questions
Both are angles between the line of sight and the horizontal through the observer’s eye. The angle of elevation is used when the point viewed is above eye level, so we raise our head. The angle of depression is used when the point viewed is below eye level, so we lower our head, for example looking down from a lighthouse at a ship.
The horizontal line through the observer at the top is parallel to the level ground. The line of sight cuts both lines, so the angle of depression at the top and the angle at the ground point are alternate angles, and alternate angles are equal. That is why you can mark the depression angle at the ground point and solve the triangle as usual.
Pick the ratio that contains the side you know and the side you want. If the problem involves a height and a distance along the ground, use tan. If it involves a height and a slanting length such as a rope, ladder, slide or kite string, use sin. If it involves the ground distance and the slanting length, use cos.
Add it when the angle is measured from the observer’s eyes. The triangle then starts at eye level, so it gives only the part of the object above the eyes. In NCERT Example 3 the triangle gives 28.5 m and the observer is 1.5 m tall, so the chimney is 30 m. If the object’s height is given, subtract the eye height first.
Let the height be h and one distance be x. Write one tan equation for each right triangle, for example h = β3x and h = (x + 40)/β3 in the shadow problem. Substitute one equation into the other to find x, then find h. For the same height, the distance at 30Β° is three times the distance at 60Β°.
The standing part is the vertical side, the broken part is the hypotenuse, and the distance from the foot to where the top touches the ground is the base. Find the standing part with tan and the broken part with cos, then add them. In NCERT Exercise 9.1 Q2 this gives 8/β3 + 16/β3 = 8β3 m.
The Sun’s altitude is the angle of elevation of the Sun. It equals the angle of elevation of the top of a vertical object from the tip of its shadow, so shadow length = height Γ cot ΞΈ. A lower Sun gives a longer shadow. If a pole and its shadow are equal in length, tan ΞΈ = 1 and the altitude is 45Β°.
