Class 10 Mathematics - NCERT Mathematics
NCERT Solutions for Class 10 Maths Chapter 9: Some Applications of Trigonometry
Chapter 9 deals with the practical applications of trigonometry in finding heights and distances. Students learn to use trigonometric ratios to calculate the height of a tower, the width of a river, the distance of a ship from a lighthouse, and similar real-world problems. The key concepts include line of sight, angle of elevation, angle of depression, and the relationship between them using alternate interior angles. This chapter is part of the Trigonometry unit which carries 12 marks in the CBSE Class 10 board exam 2027, and this chapter typically contributes 4-5 marks.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 9 Some Applications of Trigonometry has only 1 exercise - Exercise 9.1 with 16 questions. All 16 questions are included below. This chapter focuses entirely on word problems involving heights and distances using trigonometric ratios. Students must be thorough with trigonometric values of standard angles (30, 45, 60 degrees) from Chapter 8 before attempting these problems.
Table of Contents
9.1Exercise 9.1 - Heights and Distances
Q1A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30 degrees.
Let AB be the vertical pole of height h and AC be the rope of length 20 m. The angle made by the rope with the ground is angle ACB = 30 degrees.
In right triangle ABC:
sin 30 = AB/AC = h/20
1/2 = h/20
h = 20/2 = 10
The height of the pole is 10 m.
Q2A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30 degrees 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.
Let AB be the original tree. It breaks at point C. The broken part CB bends to touch the ground at D.
So CD = CB (broken part), angle CDA = 30 degrees, AD = 8 m.
In right triangle ACD:
tan 30 = AC/AD
1/sqrt(3) = AC/8
AC = 8/sqrt(3) = 8sqrt(3)/3 m
Also: cos 30 = AD/CD
sqrt(3)/2 = 8/CD
CD = 16/sqrt(3) = 16sqrt(3)/3 m
Total height = AC + CD = 8sqrt(3)/3 + 16sqrt(3)/3 = 24sqrt(3)/3 = 8sqrt(3)
The height of the tree is 8sqrt(3) m (approximately 13.86 m).
Q3A 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 degrees 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 degrees to the ground. What should be the length of the slide in each case?
Slide 1 (younger children):
Height = 1.5 m, angle of inclination = 30 degrees.
sin 30 = height/length of slide
1/2 = 1.5/l
l = 1.5 x 2 = 3
Length of slide for younger children = 3 m.
Slide 2 (elder children):
Height = 3 m, angle of inclination = 60 degrees.
sin 60 = height/length of slide
sqrt(3)/2 = 3/l
l = 6/sqrt(3) = 6sqrt(3)/3 = 2sqrt(3)
Length of slide for elder children = 2sqrt(3) m (approximately 3.46 m).
Q4The 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 degrees. Find the height of the tower.
Let AB be the tower of height h. Point C is on the ground, 30 m from the foot.
In right triangle ABC:
tan 30 = AB/BC = h/30
1/sqrt(3) = h/30
h = 30/sqrt(3) = 30sqrt(3)/3 = 10sqrt(3)
The height of the tower is 10sqrt(3) m (approximately 17.32 m).
Q5A 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 degrees. Find the length of the string, assuming that there is no slack in the string.
Let A be the position of the kite at height 60 m. B is the point on the ground where the string is tied. AB is the string.
In right triangle:
sin 60 = height/length of string = 60/l
sqrt(3)/2 = 60/l
l = 120/sqrt(3) = 120sqrt(3)/3 = 40sqrt(3)
The length of the string is 40sqrt(3) m (approximately 69.28 m).
Q6A 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 degrees to 60 degrees as he walks towards the building. Find the distance he walked towards the building.
Height of building above boy's eyes = 30 - 1.5 = 28.5 m.
Let the initial distance from the building be d1 and final distance be d2.
From initial position: tan 30 = 28.5/d1
1/sqrt(3) = 28.5/d1
d1 = 28.5sqrt(3) m
From final position: tan 60 = 28.5/d2
sqrt(3) = 28.5/d2
d2 = 28.5/sqrt(3) = 28.5sqrt(3)/3 = 9.5sqrt(3) m
Distance walked = d1 - d2 = 28.5sqrt(3) - 9.5sqrt(3) = 19sqrt(3)
The boy walked 19sqrt(3) m (approximately 32.91 m) towards the building.
Q7From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower on the top of a 20 m high building are 45 degrees and 60 degrees respectively. Find the height of the tower.
Let AB = building = 20 m, BC = tower = h m. Point P is on the ground.
Angle of elevation of bottom of tower (B) = 45 degrees.
Angle of elevation of top of tower (C) = 60 degrees.
In right triangle ABP: tan 45 = AB/BP
1 = 20/BP
BP = 20 m
In right triangle ACP: tan 60 = AC/BP = (20 + h)/20
sqrt(3) = (20 + h)/20
20sqrt(3) = 20 + h
h = 20sqrt(3) - 20 = 20(sqrt(3) - 1)
The height of the tower is 20(sqrt(3) - 1) m (approximately 14.64 m).
Q8A 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 degrees and from the same point the angle of elevation of the top of the pedestal is 45 degrees. Find the height of the pedestal.
Let AB = pedestal = h m, BC = statue = 1.6 m. Point P is on the ground.
Angle of elevation of top of pedestal (B) = 45 degrees.
Angle of elevation of top of statue (C) = 60 degrees.
In right triangle ABP: tan 45 = h/BP
1 = h/BP, so BP = h
In right triangle ACP: tan 60 = (h + 1.6)/BP = (h + 1.6)/h
sqrt(3) = (h + 1.6)/h
sqrt(3) x h = h + 1.6
h(sqrt(3) - 1) = 1.6
h = 1.6/(sqrt(3) - 1)
Rationalising: h = 1.6(sqrt(3) + 1)/((sqrt(3) - 1)(sqrt(3) + 1)) = 1.6(sqrt(3) + 1)/(3 - 1) = 1.6(sqrt(3) + 1)/2
h = 0.8(sqrt(3) + 1)
The height of the pedestal is 0.8(sqrt(3) + 1) m (approximately 2.19 m).
Q9The angle of elevation of the top of a building from the foot of the tower is 30 degrees and the angle of elevation of the top of the tower from the foot of the building is 60 degrees. If the tower is 50 m high, find the height of the building.
Let AB = tower = 50 m, CD = building = h m. Distance between them = BC = d.
From foot of tower (B), angle of elevation of top of building (D) = 30 degrees.
tan 30 = CD/BC = h/d
1/sqrt(3) = h/d ... (i)
From foot of building (C), angle of elevation of top of tower (A) = 60 degrees.
tan 60 = AB/BC = 50/d
sqrt(3) = 50/d
d = 50/sqrt(3) ... (ii)
Substituting (ii) in (i):
1/sqrt(3) = h/(50/sqrt(3))
1/sqrt(3) = h x sqrt(3)/50
50/(sqrt(3) x sqrt(3)) = h
h = 50/3
The height of the building is 50/3 m (approximately 16.67 m).
Q10Two 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 degrees and 30 degrees respectively. Find the height of the poles and the distances of the point from the poles.
Let AB and CD be two poles of equal height h. The road BC = 80 m. Point P is on the road between them.
Let BP = x, then PC = 80 - x.
From P, angle of elevation of A = 60 degrees:
tan 60 = h/x
sqrt(3) = h/x
h = x sqrt(3) ... (i)
From P, angle of elevation of D = 30 degrees:
tan 30 = h/(80 - x)
1/sqrt(3) = h/(80 - x)
h = (80 - x)/sqrt(3) ... (ii)
From (i) and (ii):
x sqrt(3) = (80 - x)/sqrt(3)
3x = 80 - x
4x = 80
x = 20 m
So BP = 20 m, PC = 60 m.
h = 20sqrt(3) m
The height of each pole is 20sqrt(3) m (approximately 34.64 m). The point is 20 m from one pole and 60 m from the other.
Q11A 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 degrees. 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 degrees. Find the height of the tower and the width of the canal.
Let AB = tower = h m. C is the point on the opposite bank, D is 20 m further from C.
BC = width of canal = d. BD = d + 20.
From C: tan 60 = h/d
sqrt(3) = h/d
h = d sqrt(3) ... (i)
From D: tan 30 = h/(d + 20)
1/sqrt(3) = h/(d + 20)
h = (d + 20)/sqrt(3) ... (ii)
From (i) and (ii):
d sqrt(3) = (d + 20)/sqrt(3)
3d = d + 20
2d = 20
d = 10 m
h = 10sqrt(3) m
The height of the tower is 10sqrt(3) m (approximately 17.32 m) and the width of the canal is 10 m.
Q12From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60 degrees and the angle of depression of its foot is 45 degrees. Determine the height of the tower.
Let AB = building = 7 m, CD = tower. From the top of the building (A), angle of elevation of top of tower (D) = 60 degrees, angle of depression of foot of tower (C) = 45 degrees.
Draw AE perpendicular to CD. Then AE = BC (horizontal distance) and CE = AB = 7 m.
Angle of depression of C = 45 degrees:
tan 45 = AB/BC = 7/BC
1 = 7/BC
BC = 7 m, so AE = 7 m
Angle of elevation of D from A:
tan 60 = DE/AE
sqrt(3) = DE/7
DE = 7sqrt(3) m
Total height of tower CD = CE + DE = 7 + 7sqrt(3) = 7(1 + sqrt(3))
The height of the tower is 7(1 + sqrt(3)) m (approximately 19.12 m).
Q13As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30 degrees and 45 degrees. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Let AB = lighthouse = 75 m. Ships are at C and D. Angle of depression of C = 45 degrees, angle of depression of D = 30 degrees.
From A: tan 45 = 75/BC
1 = 75/BC
BC = 75 m
tan 30 = 75/BD
1/sqrt(3) = 75/BD
BD = 75sqrt(3) m
Distance between ships = BD - BC = 75sqrt(3) - 75 = 75(sqrt(3) - 1)
The distance between the two ships is 75(sqrt(3) - 1) m (approximately 54.90 m).
Q14A 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 degrees. After some time, the angle of elevation reduces to 30 degrees. Find the distance travelled by the balloon during the interval.
Height of balloon above girl's eyes = 88.2 - 1.2 = 87 m.
Let the initial position of balloon be at B and final position at C. Let the girl be at A.
From initial position: tan 60 = 87/d1
sqrt(3) = 87/d1
d1 = 87/sqrt(3) = 29sqrt(3) m
From final position: tan 30 = 87/d2
1/sqrt(3) = 87/d2
d2 = 87sqrt(3) m
Distance travelled by balloon = d2 - d1 = 87sqrt(3) - 29sqrt(3) = 58sqrt(3)
The balloon travelled 58sqrt(3) m (approximately 100.46 m).
Q15A 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 degrees, 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 degrees. Find the time taken by the car to reach the foot of the tower from this point.
Let AB = tower = h. Car moves from C to D in 6 seconds.
From A, angle of depression of C = 30 degrees:
tan 30 = h/BC
BC = h sqrt(3) ... (i)
Angle of depression of D = 60 degrees:
tan 60 = h/BD
BD = h/sqrt(3) = h sqrt(3)/3 ... (ii)
Distance CD = BC - BD = h sqrt(3) - h sqrt(3)/3 = h sqrt(3)(1 - 1/3) = 2h sqrt(3)/3
Car covers CD = 2h sqrt(3)/3 in 6 seconds.
Speed = (2h sqrt(3)/3)/6 = h sqrt(3)/9
Time to cover BD = h sqrt(3)/3:
Time = (h sqrt(3)/3)/(h sqrt(3)/9) = (h sqrt(3)/3) x (9/(h sqrt(3))) = 9/3 = 3 seconds
The car takes 3 seconds to reach the foot of the tower from point D.
Q16The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Let AB = tower = h. Points C and D are at distances 4 m and 9 m from the base B respectively.
Angles of elevation are complementary, so if one is theta, the other is (90 - theta).
From C: tan theta = h/4 ... (i)
From D: tan(90 - theta) = h/9
cot theta = h/9
1/tan theta = h/9 ... (ii)
Multiplying (i) and (ii):
tan theta x (1/tan theta) = (h/4) x (h/9)
1 = h squared/36
h squared = 36
h = 6 m (taking positive value)
Hence, the height of the tower is 6 m. Proved.
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From the top of a 15 m high building, the angle of elevation of the top of a tower is 60 degrees and the angle of depression of the foot of the tower is 30 degrees. Find the height of the tower.
Let building AB = 15 m, tower CD. Draw AE perpendicular to CD. CE = AB = 15 m, AE = BC.
tan 30 = 15/BC, so BC = 15sqrt(3) m. AE = 15sqrt(3) m.
tan 60 = DE/AE, so DE = 15sqrt(3) x sqrt(3) = 45 m.
Height of tower = CE + DE = 15 + 45 = 60 m.
The shadow of a tower standing on level ground is found to be 40 m longer when the Sun's altitude is 30 degrees than when it is 60 degrees. Find the height of the tower.
Let height = h. Shadow at 60 degrees = h/sqrt(3). Shadow at 30 degrees = h sqrt(3).
Difference = h sqrt(3) - h/sqrt(3) = 40.
h(sqrt(3) - 1/sqrt(3)) = 40.
h(3 - 1)/sqrt(3) = 40.
2h/sqrt(3) = 40.
h = 20sqrt(3) m.
Height of the tower = 20sqrt(3) m (approximately 34.64 m).
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60 degrees. When he moves 40 m away from the bank, the angle of elevation becomes 30 degrees. Find the height of the tree and the width of the river.
Let tree height = h, river width = d.
tan 60 = h/d, so h = d sqrt(3) ... (i)
tan 30 = h/(d + 40), so h = (d + 40)/sqrt(3) ... (ii)
From (i) and (ii): d sqrt(3) = (d + 40)/sqrt(3)
3d = d + 40, so 2d = 40, d = 20 m.
h = 20sqrt(3) m.
Height = 20sqrt(3) m (approximately 34.64 m), width of river = 20 m.
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 degrees, 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 degrees. Find the time taken by the car to reach the foot of the tower from this point.
Let tower height = h. At 30 degrees: BC = h sqrt(3). At 60 degrees: BD = h/sqrt(3).
Distance CD = h sqrt(3) - h/sqrt(3) = 2h sqrt(3)/3, covered in 6 seconds.
Speed = 2h sqrt(3)/(3 x 6) = h sqrt(3)/9.
Time to cover BD = (h/sqrt(3))/(h sqrt(3)/9) = (h/sqrt(3)) x (9/(h sqrt(3))) = 9/3 = 3 seconds.
From a point on the ground, the angle of elevation of the bottom and top of a transmission tower on the top of a 20 m high building are 45 degrees and 60 degrees respectively. Find the height of the transmission tower.
Let building = 20 m, tower = h m, distance from point to building = d.
tan 45 = 20/d, so d = 20 m.
tan 60 = (20 + h)/20, so sqrt(3) = (20 + h)/20.
20 + h = 20sqrt(3), h = 20sqrt(3) - 20 = 20(sqrt(3) - 1).
Height of tower = 20(sqrt(3) - 1) m (approximately 14.64 m).
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 degrees and 30 degrees respectively. Find the height of the poles and the distances of the point from the poles.
Let height = h, distances from point = x and (80 - x).
tan 60 = h/x, so h = x sqrt(3) ... (i)
tan 30 = h/(80 - x), so h = (80 - x)/sqrt(3) ... (ii)
From (i) and (ii): x sqrt(3) = (80 - x)/sqrt(3). 3x = 80 - x. 4x = 80. x = 20 m.
h = 20sqrt(3) m. Distances: 20 m and 60 m.
Height = 20sqrt(3) m (approximately 34.64 m). Distances = 20 m and 60 m.
An observer, 1.5 m tall, is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45 degrees. What is the height of the chimney?
The observer is 1.5 m tall and stands 28.5 m from the chimney. The angle of elevation from the observer's eyes to the top of the chimney is 45 degrees.
Height of chimney above the observer's eyes: tan 45 = h/28.5, so 1 = h/28.5, giving h = 28.5 m.
Total height of chimney = height above eyes + observer's height = 28.5 + 1.5 = 30 m.
From the top of a 7 m high building, the angle of elevation of the top of a tower is 60 degrees and the angle of depression of its foot is 45 degrees. Find the height of the tower. (Use sqrt(3) = 1.732)
Building = 7 m. tan 45 = 7/d, so d = 7 m.
tan 60 = DE/7, so DE = 7sqrt(3).
Total height = 7 + 7sqrt(3) = 7(1 + sqrt(3)) = 7(1 + 1.732) = 7 x 2.732 = 19.124 m.
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 45 degrees. Find the height of the tower.
Let the height of the tower be h metres. The point of observation is 30 m away from the foot of the tower, and the angle of elevation is 45 degrees.
Using tan 45 = h/30:
1 = h/30
h = 30
The height of the tower is 30 m.
A statue 1.46 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 degrees and from the same point, the angle of elevation of the top of the pedestal is 45 degrees. Find the height of the pedestal.
Let pedestal = h. tan 45 = h/d, so d = h.
tan 60 = (h + 1.46)/h. sqrt(3) = (h + 1.46)/h.
h sqrt(3) = h + 1.46. h(sqrt(3) - 1) = 1.46.
h = 1.46/(sqrt(3) - 1) = 1.46(sqrt(3) + 1)/2 = 0.73(sqrt(3) + 1).
h = 0.73(sqrt(3) + 1) m (approximately 1.99 m).
A boat is being rowed away from a cliff 150 m high. From the top of the cliff, the angle of depression of the boat changes from 60 degrees to 45 degrees in 2 minutes. Find the speed of the boat.
At 60 degrees: tan 60 = 150/d1. d1 = 150/sqrt(3) = 50sqrt(3) m.
At 45 degrees: tan 45 = 150/d2. d2 = 150 m.
Distance covered = 150 - 50sqrt(3) = 50(3 - sqrt(3)) m in 2 minutes.
Speed = 50(3 - sqrt(3))/2 = 25(3 - sqrt(3)) m/min (approximately 31.70 m/min).
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60 degrees and the angle of depression of its foot is 30 degrees. Determine the height of the tower.
Building = 7 m. tan 30 = 7/d. d = 7sqrt(3) m.
tan 60 = DE/(7sqrt(3)). DE = 7sqrt(3) x sqrt(3) = 21 m.
Height of tower = 7 + 21 = 28 m.
A kite is flying at a height of 30 m from the ground. The length of string from the kite to the ground is 60 m. Assuming that there is no slack in the string, the angle of elevation of the kite at the ground is:
The kite is flying at a height of 30 m from the ground. The length of the string from the kite to the ground is 60 m. We need to find the angle of elevation.
Using sin theta = height/string length = 30/60 = 1/2.
Since sin 30 = 1/2, the angle of elevation theta = 30 degrees.
The angle of depression of a car parked on the road from the top of a 150 m high tower is 30 degrees. Find the distance of the car from the tower.
The tower is 150 m high and the angle of depression of the car from the top of the tower is 30 degrees. Since the angle of depression equals the angle of elevation (alternate interior angles), the angle of elevation from the car to the top is also 30 degrees.
Using tan 30 = 150/d:
1/sqrt(3) = 150/d
d = 150sqrt(3)
The distance of the car from the tower is 150sqrt(3) m (approximately 259.81 m).
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): The angle of elevation of the top of a tower from a point 20 m away from its base is 45 degrees. The height of the tower is 20 m.
Reason (R): tan 45 = perpendicular/base, and tan 45 = 1.
Answer: (A)
Both are true. tan 45 = h/20, so 1 = h/20, giving h = 20 m. The reason correctly explains why the height equals the distance.
AR2
Assertion (A): If the angle of depression from the top of a building to a point on the ground is 30 degrees, then the angle of elevation from that point to the top of the building is also 30 degrees.
Reason (R): Angle of depression and angle of elevation are alternate interior angles formed by the line of sight and two parallel horizontal lines.
Answer: (A)
Both are true. The horizontal from the top of the building and the horizontal at ground level are parallel. The line of sight acts as a transversal, making angle of depression and angle of elevation alternate interior angles, hence equal.
AR3
Assertion (A): If the height of a tower is sqrt(3) times the length of its shadow, then the angle of elevation of the Sun is 60 degrees.
Reason (R): tan 60 = sqrt(3).
Answer: (A)
Both are true. If height = sqrt(3) x shadow, then tan theta = height/shadow = sqrt(3). Since tan 60 = sqrt(3), the angle is 60 degrees. The reason correctly explains the assertion.
AR4
Assertion (A): A person standing at a height of 10 m observes the angle of depression of a boat as 45 degrees. The boat is 10 m away from the base of the tower.
Reason (R): tan 30 = 1/sqrt(3).
Answer: (C)
Assertion is TRUE: tan 45 = 10/d, so d = 10 m. Reason is TRUE as a fact (tan 30 = 1/sqrt(3)), but it does NOT explain the assertion which uses tan 45, not tan 30.
AR5
Assertion (A): The shadow of a tower becomes sqrt(3) times longer when the angle of elevation of the Sun changes from 60 degrees to 30 degrees.
Reason (R): The ratio of shadows at 30 degrees and 60 degrees is tan 60/tan 30 = sqrt(3)/(1/sqrt(3)) = 3.
Answer: (D)
Assertion is FALSE: shadow at 60 = h/sqrt(3), shadow at 30 = h sqrt(3). Ratio = h sqrt(3)/(h/sqrt(3)) = 3, not sqrt(3). The shadow becomes 3 times longer. Reason is TRUE as the calculation is correct.
AR6
Assertion (A): From the top of a 60 m high lighthouse, the angles of depression of two ships are 30 degrees and 60 degrees. The distance between the ships is 40sqrt(3) m.
Reason (R): The distance between the ships = 60sqrt(3) - 60/sqrt(3) = 60sqrt(3) - 20sqrt(3) = 40sqrt(3) m.
Answer: (A)
Both are true. Ship 1 at 60 degrees: d1 = 60/sqrt(3) = 20sqrt(3) m. Ship 2 at 30 degrees: d2 = 60sqrt(3) m. Distance = 60sqrt(3) - 20sqrt(3) = 40sqrt(3) m. The reason correctly calculates and explains the assertion.
Case Study Questions
A lighthouse stands on the edge of a cliff. From the top of the lighthouse, which is 80 m above sea level, a coastguard observes two ships approaching. Ship A is directly ahead and Ship B is behind Ship A in the same line. The angles of depression of Ship A and Ship B are 60 degrees and 30 degrees respectively.
(i) What is the distance of Ship A from the base of the lighthouse? (1 mark)
tan 60 = 80/d(A). sqrt(3) = 80/d(A). d(A) = 80/sqrt(3) = 80sqrt(3)/3.
Distance of Ship A = 80sqrt(3)/3 m (approximately 46.19 m).
(ii) What is the distance of Ship B from the base of the lighthouse? (1 mark)
tan 30 = 80/d(B). 1/sqrt(3) = 80/d(B). d(B) = 80sqrt(3).
Distance of Ship B = 80sqrt(3) m (approximately 138.56 m).
(iii)(a) Find the distance between the two ships. (2 marks)
Distance = d(B) - d(A) = 80sqrt(3) - 80sqrt(3)/3 = (240sqrt(3) - 80sqrt(3))/3 = 160sqrt(3)/3.
Distance between ships = 160sqrt(3)/3 m (approximately 92.38 m).
(iii)(b) If Ship A moves to the position where the angle of depression becomes 45 degrees, how far has it moved from its original position? (2 marks)
New distance of Ship A: tan 45 = 80/d. d = 80 m.
Original distance = 80sqrt(3)/3 m.
Since 80 > 80sqrt(3)/3 (as 80 > 46.19), Ship A moved away.
Distance moved = 80 - 80sqrt(3)/3 = 80(1 - sqrt(3)/3) = 80(3 - sqrt(3))/3.
Ship A moved 80(3 - sqrt(3))/3 m (approximately 33.81 m) away from the lighthouse.
A park has an observation tower AB of height 30 m. Two visitors P and Q are standing on either side of the tower. The angle of elevation of the top of the tower from P is 30 degrees and from Q is 45 degrees. Both P and Q are on the same horizontal level as the foot of the tower.
(i) Find the distance of point P from the base of the tower. (1 mark)
tan 30 = 30/BP. 1/sqrt(3) = 30/BP. BP = 30sqrt(3).
Distance of P from the tower = 30sqrt(3) m (approximately 51.96 m).
(ii) Find the distance of point Q from the base of the tower. (1 mark)
tan 45 = 30/BQ. 1 = 30/BQ. BQ = 30.
Distance of Q from the tower = 30 m.
(iii)(a) Find the distance between P and Q. (2 marks)
P and Q are on opposite sides of the tower.
PQ = BP + BQ = 30sqrt(3) + 30 = 30(sqrt(3) + 1).
Distance between P and Q = 30(sqrt(3) + 1) m (approximately 81.96 m).
(iii)(b) If the height of the tower is doubled to 60 m, what would be the new distance of P from the base? (2 marks)
tan 30 = 60/BP(new). 1/sqrt(3) = 60/BP(new). BP(new) = 60sqrt(3).
New distance of P = 60sqrt(3) m (approximately 103.92 m), which is also double the original distance.
MCQ Practice
MCQ1. The angle of elevation of the top of a tower from a point on the ground, 30 m away from the foot of the tower, is 30 degrees. The height of the tower is:
(A) 10sqrt(3) m
(B) 30sqrt(3) m
(C) 30 m
(D) 10 m
Answer: (A) - tan 30 = h/30. 1/sqrt(3) = h/30. h = 30/sqrt(3) = 10sqrt(3) m.
MCQ2. If the height of a building and the distance of the point of observation from its foot, both are increased by 10%, then the angle of elevation of its top:
(A) increases
(B) decreases
(C) remains unchanged
(D) depends on original values
Answer: (C) - tan theta = h/d. If both increase by 10%: tan theta = 1.1h/1.1d = h/d. The ratio stays the same, so the angle remains unchanged.
MCQ3. A pole 6 m high casts a shadow 2sqrt(3) m long on the ground, then the Sun's elevation is:
(A) 30 degrees
(B) 45 degrees
(C) 60 degrees
(D) 90 degrees
Answer: (C) - tan theta = 6/(2sqrt(3)) = 3/sqrt(3) = sqrt(3). Since tan 60 = sqrt(3), the Sun's elevation is 60 degrees.
MCQ4. The angle of depression of a car standing on the ground from the top of a 75 m tower is 30 degrees. The distance of the car from the base of the tower is:
(A) 25sqrt(3) m
(B) 75sqrt(3) m
(C) 150 m
(D) 50sqrt(3) m
Answer: (B) - Angle of depression = 30 degrees = angle of elevation. tan 30 = 75/d. 1/sqrt(3) = 75/d. d = 75sqrt(3) m.
MCQ5. A ladder 15 m long leans against a wall making an angle of 60 degrees with the ground. The height of the wall up to the point where the ladder touches is:
(A) 15sqrt(3)/2 m
(B) 15/2 m
(C) 15 m
(D) 15sqrt(3) m
Answer: (A) - sin 60 = height/15. sqrt(3)/2 = height/15. height = 15sqrt(3)/2 m.
MCQ6. If a tower is 30 m high and casts a shadow of 10sqrt(3) m, then the angle of elevation of the Sun is:
(A) 30 degrees
(B) 45 degrees
(C) 60 degrees
(D) 75 degrees
Answer: (C) - tan theta = 30/(10sqrt(3)) = 3/sqrt(3) = sqrt(3). tan 60 = sqrt(3). Angle = 60 degrees.
MCQ7. From the top of a cliff 20 m high, the angle of elevation of the top of a tower is found to be the same as the angle of depression of the foot of the tower. The height of the tower is:
(A) 20 m
(B) 40 m
(C) 60 m
(D) 80 m
Answer: (B) - Let angle of depression = angle of elevation = theta. tan theta = 20/d, so d = 20/tan theta. Height above cliff = d x tan theta = 20. Total height = 20 + 20 = 40 m.
MCQ8. The angles of elevation of the top of a tower from two points at distances a and b from the base (a > b) are complementary. The height of the tower is:
(A) sqrt(a + b)
(B) sqrt(a - b)
(C) sqrt(ab)
(D) a/b
Answer: (C) - tan theta = h/b and tan(90 - theta) = h/a. So cot theta = h/a, i.e., 1/tan theta = h/a. Multiplying: 1 = h^2/(ab). h^2 = ab. h = sqrt(ab).
MCQ9. A kite is flying at a height of 30 m from the ground. The length of the string from the kite to the ground is 60 m. The angle of elevation of the kite from the ground is:
(A) 30 degrees
(B) 45 degrees
(C) 60 degrees
(D) 90 degrees
Answer: (A) - sin theta = 30/60 = 1/2. Since sin 30 = 1/2, the angle of elevation is 30 degrees.
MCQ10. If the angle of elevation of the Sun changes from 45 degrees to 60 degrees, the length of the shadow of a pillar decreases by 10 m. The height of the pillar is:
(A) 5(3 + sqrt(3)) m
(B) 5(3 - sqrt(3)) m
(C) 10(sqrt(3) + 1) m
(D) 10(sqrt(3) - 1) m
Answer: (A) - Shadow at 45 degrees = h (since tan 45 = 1). Shadow at 60 degrees = h/sqrt(3). Decrease: h - h/sqrt(3) = 10. h(sqrt(3) - 1)/sqrt(3) = 10. h = 10sqrt(3)/(sqrt(3) - 1). Rationalising: h = 10sqrt(3)(sqrt(3) + 1)/2 = 5(3 + sqrt(3)) m, approximately 23.66 m.
MCQ11. If the height and the length of the shadow of a man are equal, then the angle of elevation of the Sun is:
(A) 30 degrees
(B) 45 degrees
(C) 60 degrees
(D) 90 degrees
Answer: (B) - tan theta = height/shadow = h/h = 1. Since tan 45 = 1, the angle of elevation is 45 degrees.
MCQ12. The angle of elevation of the top of a tower is 30 degrees. If the height of the tower is doubled, then the angle of elevation of its top will:
(A) also get doubled
(B) will not change
(C) will become less than 60 degrees
(D) will become 60 degrees
Answer: (C) - Original: tan 30 = h/d, so d = h sqrt(3). New: tan theta = 2h/d = 2h/(h sqrt(3)) = 2/sqrt(3). Now tan 60 = sqrt(3) = 1.732 and 2/sqrt(3) = 1.155. Since 1.155 < 1.732, theta < 60 degrees. So the new angle is less than 60 degrees.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
A pole 6 m high casts a shadow 2sqrt(3) m long. What is the angle of elevation of the Sun?
tan theta = 6/(2sqrt(3)) = sqrt(3). So theta = 60 degrees.
If the angle of depression from the top of a tower to a point on the ground is 30 degrees, what is the angle of elevation from that point to the top of the tower?
The angle of elevation is also 30 degrees, because angle of depression and angle of elevation are alternate interior angles (the horizontal lines are parallel).
A kite is flying at a height of 30 m. The string attached to the kite is 60 m long. Find the angle the string makes with the ground.
sin theta = 30/60 = 1/2. So theta = 30 degrees.
What is the angle of elevation of the Sun when the height of a vertical pole is equal to the length of its shadow?
tan theta = height/shadow = 1. So theta = 45 degrees.
The angle of elevation of the top of a tower from a point 10 m away from its base is 60 degrees. Find the height of the tower.
tan 60 = h/10. sqrt(3) = h/10. h = 10sqrt(3) m (approximately 17.32 m).
2-3 Mark Questions
The shadow of a tower standing on level ground is found to be 40 m longer when the Sun's altitude is 30 degrees than when it is 60 degrees. Find the height of the tower.
Let height = h.
Shadow at 60 degrees: tan 60 = h/s1, s1 = h/sqrt(3).
Shadow at 30 degrees: tan 30 = h/s2, s2 = h sqrt(3).
Difference: s2 - s1 = h sqrt(3) - h/sqrt(3) = 40
h(sqrt(3) - 1/sqrt(3)) = 40
h(3 - 1)/sqrt(3) = 40
2h/sqrt(3) = 40
h = 20sqrt(3)
Height of the tower = 20sqrt(3) m (approximately 34.64 m).
An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from the observer's eyes is 45 degrees. What is the height of the chimney?
Height above eyes: tan 45 = h/28.5, so h = 28.5 m.
Total height of chimney = 28.5 + 1.5 = 30 m.
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60 degrees and the angle of depression of its foot is 45 degrees. Determine the height of the tower.
tan 45 = 7/d, so d = 7 m.
tan 60 = DE/7, so DE = 7sqrt(3) m.
Total height = 7 + 7sqrt(3) = 7(1 + sqrt(3)).
Height of the tower = 7(1 + sqrt(3)) m (approximately 19.12 m).
3-4 Mark Questions
A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank is 60 degrees. When he moves 40 m away from the bank, the angle becomes 30 degrees. Find the height of the tree and width of the river.
Let tree height = h, river width = d.
tan 60 = h/d, so h = d sqrt(3) ... (i)
tan 30 = h/(d + 40), so h = (d + 40)/sqrt(3) ... (ii)
From (i) and (ii): d sqrt(3) = (d + 40)/sqrt(3)
3d = d + 40
2d = 40, d = 20 m
h = 20sqrt(3) m
Height = 20sqrt(3) m (approximately 34.64 m), width of river = 20 m.
Two poles of equal heights are standing opposite each other on either side of a road 80 m wide. From a point between them, the angles of elevation of the tops are 60 degrees and 30 degrees. Find the heights and distances.
Let height = h, distances = x and (80 - x).
tan 60 = h/x, so h = x sqrt(3) ... (i)
tan 30 = h/(80 - x), so h = (80 - x)/sqrt(3) ... (ii)
From (i) and (ii): x sqrt(3) = (80 - x)/sqrt(3)
3x = 80 - x, 4x = 80, x = 20 m
h = 20sqrt(3) m
Height = 20sqrt(3) m (approximately 34.64 m). Distances: 20 m and 60 m.
A straight highway leads to the foot of a tower. A man standing at the top observes a car at an angle of depression of 30 degrees, which is approaching at uniform speed. Six seconds later, the angle of depression is 60 degrees. Find the time taken by the car to reach the foot of the tower.
Let tower = h. At 30 degrees: BC = h sqrt(3). At 60 degrees: BD = h/sqrt(3).
CD = h sqrt(3) - h/sqrt(3) = 2h sqrt(3)/3, covered in 6 sec.
Speed = h sqrt(3)/9.
Time for BD = (h/sqrt(3)) / (h sqrt(3)/9) = 9/3 = 3 seconds.
Key Concepts
Important Formulas
Study Plan - Some Applications of Trigonometry
Understanding Angles of Elevation and Depression
Read NCERT theory. Understand line of sight, angle of elevation, and angle of depression with diagrams. Learn how to identify the right triangle in any problem. Practice drawing figures for Q1-Q5.
Basic Height and Distance Problems
Solve Exercise 9.1 Q1-Q8. Focus on identifying which trig ratio to use (sin, cos, or tan) based on the given information. Most problems use tan.
Two-Angle Problems and Moving Objects
Solve Exercise 9.1 Q9-Q16. These involve two angles and require setting up two equations. Practice the pattern: two equations, eliminate the common variable.
Previous Year Questions and MCQs
Solve all previous year questions. Practice MCQs and assertion-reason questions. Time yourself - each 3-mark problem should take under 5 minutes.
Revision and Case Studies
Solve case study questions. Revise all trig values (sin, cos, tan for 30, 45, 60). Practice the complementary angle property. Solve 5 mixed problems without looking at solutions.
Common Mistakes to Avoid
- ✗Confusing angle of elevation with angle of depression - elevation is when looking UP from horizontal, depression is when looking DOWN from horizontal
- ✗Forgetting that the angle of depression from top equals the angle of elevation from bottom (alternate interior angles) - this is the most tested concept
- ✗Using the wrong trig ratio - tan is used when height and distance are given (most common), sin when height and hypotenuse are involved, cos when base and hypotenuse are involved
- ✗Not drawing the figure correctly - always mark the right angle, label all known values, and identify which side is perpendicular and which is base before applying any formula
- ✗Mixing up tan 30 = 1/sqrt(3) and tan 60 = sqrt(3) - remember tan 30 < 1 (small angle, small ratio) and tan 60 > 1 (large angle, large ratio)
Tips for Board Exam
- ✓Always draw a clear, labelled diagram first - examiners award marks for the diagram itself (typically 1 mark out of 3-4). Mark the right angle, label heights, distances, and angles clearly.
- ✓Memorise the trig values table for 30, 45, and 60 degrees before the exam - every problem in this chapter uses these values. Write them on the margin as soon as the exam starts.
- ✓For two-angle problems (the most common board exam pattern), always set up two equations using tan and then divide or subtract to eliminate the unknown distance. This pattern appears in Q9-Q16 and in most PYQs.
- ✓In problems involving both angle of elevation and depression, draw a horizontal line from the observer and use the fact that angle of depression = angle of elevation (alternate interior angles). State this property explicitly for full marks.
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Start Some Applications of Trigonometry - Choose Your ModeAbout Some Applications of Trigonometry - Class 10 Mathematics
What Are Heights and Distances?
Heights and distances is the branch of trigonometry that deals with measuring the height of objects (towers, buildings, mountains) and distances (width of rivers, distance of ships) without actually measuring them physically. The idea is simple: if you know one side and one angle of a right triangle, you can find the other sides using trigonometric ratios. This chapter applies the trigonometric ratios learned in Chapter 8 (Introduction to Trigonometry) to solve real-world problems.
In everyday life, surveyors use this technique to measure the height of buildings, engineers calculate distances across valleys, and navigators determine the distance of ships from the shore. The problems in this chapter are modelled on these real scenarios.
Angle of Elevation and Depression
The line of sight is the straight line drawn from the eye of the observer to the point being viewed. When the object being viewed is above the horizontal level of the observer's eyes, the angle between the line of sight and the horizontal is called the angle of elevation. When the object is below the horizontal level, this angle is called the angle of depression.
A critical property tested in board exams: the angle of depression from point A to point B is always equal to the angle of elevation from point B to point A. This is because the horizontal lines at both points are parallel, and the line of sight acts as a transversal, making these alternate interior angles. Students must state this property explicitly when solving problems involving angle of depression.
How to Solve Heights and Distances Problems
Every heights and distances problem follows a standard approach:
Step 1: Draw the diagram. Read the problem carefully and draw a rough figure. Mark the right angle (it is always at the base of the vertical object). Label all known quantities (heights, distances, angles).
Step 2: Identify the right triangle. In every problem, there is at least one right triangle. Identify which side is the perpendicular (height), which is the base (horizontal distance), and which is the hypotenuse (slant distance like a rope or ladder).
Step 3: Choose the correct trig ratio. If the problem gives height and asks for distance (or vice versa), use tan. If it involves a rope, string, or ladder (hypotenuse), use sin or cos. The rule: tan = perpendicular/base, sin = perpendicular/hypotenuse, cos = base/hypotenuse.
Step 4: Substitute and solve. Use the standard values: tan 30 = 1/sqrt(3), tan 45 = 1, tan 60 = sqrt(3), sin 30 = 1/2, sin 60 = sqrt(3)/2, cos 30 = sqrt(3)/2, cos 60 = 1/2. Solve the resulting equation.
Step 5: Rationalise if needed. If the answer has sqrt(3) in the denominator, rationalise by multiplying numerator and denominator by sqrt(3).
Common Problem Types
Type 1: Single angle problems. Given one angle and one measurement, find the unknown. Examples: finding height of a tower given distance and angle of elevation, finding length of a rope given height and angle. These are straightforward 1-2 mark problems.
Type 2: Two-angle problems from the same side. An observer changes position (walks towards or away from the object) and the angle changes. This gives two equations with two unknowns. Solve by dividing or subtracting the equations. Examples: shadow problems, moving car problems. These are 3-4 mark problems.
Type 3: Angle of elevation and depression combined. From the top of a building, the observer sees the top of another building at an angle of elevation and its foot at an angle of depression. Draw a horizontal line from the observer and use both angles. These are the most frequently asked 3-4 mark problems in board exams.
Type 4: Two objects on opposite sides. Two poles or buildings on opposite sides of a road, with angles measured from a point between them. Set up two equations using the heights and distances on each side.
Type 5: Complementary angle problems. When two angles of elevation are complementary (sum = 90 degrees), use the property tan theta x tan(90 - theta) = 1, which gives height = sqrt(product of distances). This is a favourite for MCQs.
How Tested in Board Exams
This chapter carries 4-5 marks in the CBSE board exam as part of the 12-mark Trigonometry unit. The typical pattern is:
1 MCQ (1 mark): Usually a simple angle of elevation calculation (shadow problem, pole height) or a conceptual question about angles.
1 Short answer (2-3 marks): A straightforward heights and distances problem, often involving a single diagram with one or two angles.
1 Long answer (3-4 marks): A complex problem combining angle of elevation and depression, or a moving object problem. These require two equations and careful algebraic manipulation.
Case study questions from this chapter have appeared since CBSE 2024, typically involving a real-world scenario (lighthouse, park tower) with 3-4 sub-questions of increasing difficulty.
The most repeated problem types in CBSE 2019-2026 are: building-to-tower problems (angle of elevation + depression), shadow problems (Sun's angle changing), and moving object problems (car approaching tower).
Tips for Full Marks
- Always draw the diagram first and label everything. The diagram alone carries marks.
- Write the trig ratio formula before substituting values. For example, write "tan 60 = height/distance" before writing "sqrt(3) = h/20".
- State the alternate interior angles property when using angle of depression.
- Show rationalisation steps clearly when sqrt(3) appears in the denominator.
- Write the final answer with proper units (m or cm) and box or highlight it.
- For two-equation problems, clearly number the equations and show the substitution step.
Connection to Other Chapters
This chapter directly builds on Chapter 8 (Introduction to Trigonometry) which taught the basic trig ratios and their values for specific angles. Without knowing the values of sin, cos, and tan for 30, 45, and 60 degrees, students cannot solve any problem in this chapter. The algebraic manipulation required (rationalisation, solving simultaneous equations) connects to Chapter 1 (Real Numbers) and Chapter 3 (Pair of Linear Equations). The geometry of right triangles connects to Chapter 6 (Triangles). In real-world applications, this chapter connects to coordinate geometry (Chapter 7) when determining positions of objects.
Frequently Asked Questions
How many exercises are in NCERT Class 10 Maths Chapter 9 for CBSE 2026-27?
Chapter 9 Some Applications of Trigonometry has only 1 exercise - Exercise 9.1 with 16 questions. All questions are based on heights and distances using trigonometric ratios. There are no other exercises in this chapter.
What is the difference between angle of elevation and angle of depression?
Angle of elevation is the angle between the horizontal and the line of sight when looking UP at an object above the observer. Angle of depression is the angle between the horizontal and the line of sight when looking DOWN at an object below the observer. A key property: the angle of depression from A to B equals the angle of elevation from B to A (alternate interior angles).
Which trigonometric ratio is most commonly used in Chapter 9?
tan theta = perpendicular/base is the most commonly used ratio because most problems involve finding heights (perpendicular) and horizontal distances (base). sin and cos are used only when the hypotenuse (rope, string, ladder, wire) is involved in the problem.
How many marks does Chapter 9 carry in the CBSE board exam?
Chapter 9 is part of the Trigonometry unit which carries 12 marks total (combined with Chapter 8). Chapter 9 alone typically contributes 4-5 marks, usually as one MCQ (1 mark) and one long answer (3-4 marks) on heights and distances.
What are the most important trig values to memorise for this chapter?
You must memorise: tan 30 = 1/sqrt(3), tan 45 = 1, tan 60 = sqrt(3), sin 30 = 1/2, sin 60 = sqrt(3)/2, cos 30 = sqrt(3)/2, cos 60 = 1/2. Also remember: sqrt(3) is approximately 1.732. These values are used in every single problem.
What is the complementary angle property used in this chapter?
If two angles of elevation from two different points are complementary (their sum is 90 degrees), then the height of the object equals the square root of the product of the two distances: h = sqrt(a x b), where a and b are the distances. This comes from tan theta x tan(90 - theta) = tan theta x cot theta = 1. This property is used in Q16 of Exercise 9.1.
Do I need to draw diagrams in the board exam for this chapter?
Yes, drawing a clear, labelled diagram is essential. Examiners typically allocate 1 mark (out of 3-4) just for the diagram. Always mark the right angle, label the height, distance, and angles. Even if your calculation has a minor error, a correct diagram can earn you partial marks.
What types of problems from Chapter 9 are most frequently asked in CBSE board exams?
The most frequently asked types are: (1) Building-to-tower problems combining angle of elevation and depression (asked in 2020, 2022, 2024, 2026), (2) Shadow problems where the Sun's angle changes (asked in 2025), (3) Moving object problems like a car approaching a tower (asked in 2024), (4) Two-pole problems with angles from a point between them (asked in 2023). Two-angle problems requiring simultaneous equations are the most common pattern.