Heights and Distances Class 10 OP Malhotra Exe-17 ICSE Maths Solutions

Heights and Distances Class 10 OP Malhotra Exe-17 ICSE Maths Solutions ICSE Maths Solutions Ch-17 questions as latest prescribe guideline for upcoming exam. In this article you would learn how to solve problems on Height and Distance easily. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.

Heights and Distances Class 10 OP Malhotra Exe-17 ICSE Maths Solutions

Heights and Distances Class 10 OP Malhotra Exe-17 ICSE Maths Solutions

Board ICSE
Publications  S Chand
Subject Maths
Class 10th
Chapter-17 Heights and Distances
Writer OP Malhotra
Exe-17 Problems on Heights and Distances
Edition 2024-2025

Angle of Elevation

The angle of elevation is the angle between the horizontal and the line of sight when looking up at an object.

When looking up at an object, the angle of elevation is the angle between the horizontal line of sight and the object. For example, if you look up at a tree while walking through a forest, the angle between the ground and your line of sight to the tree is the angle of elevation

Angle of Depression

The angle of depression is the angle between the horizontal and the line of sight when looking down at an object

When looking down at an object, the angle of depression is the angle between the horizontal line of sight and the object. For example, if you look down at a ball while standing on a football field, the angle between the ground and your line of sight to the ball is the angle of depression

Exercise- 17 (Heights and Distances)

Heights and Distances Class 10 OP Malhotra Exe-17 ICSE Maths Solutions ICSE Maths Solutions Ch-17

Que-1: The angle of elevation of the top of a tower from a point at a distance of 100 metres from its foot on a horizontal plane is found to be 60°. Find the height of the tower.

Sol: Let the height of the tower AC be h metres.
Given, distance BC = 100 m, ∠ABC = 60°.
∴ tan 60° = AC/BC = h/100   [AC/BC = h/100]
⇒ h = 100 x tan 60° = (100 x √3) metre
= 100 × 1.732 = 173.2 m

Que-2: A vertical flagstaff stands on a horizontal plane. From a point distance 150 metres from its foot the angle of elevation of its top is found to be 30°; find the height of the flagstaff.

Sol:  Let AB be the vertical flagstaff of height h.
As we know that,
tanθ = Perpendicular/Base
Now, in △ABC
tan30° = AB/BC
⇒ 1/√3 = h/150
⇒ h = 150/√3
h = 50√3 = 86.6 m.

Que-3: The string of a kite is 150 m long and it makes an angle of 60° with the horizontal. Find the height of the kite from the ground.

Sol:  Let h be the height of the kite.
PB be the length of string such that PB = 150 m.
In right-angled ΔBPA,
sin 60° = h/150
⇒ √3/2 = h/150
⇒ h = (150√3)/2
⇒ h = 75√3
h = 1.732 x 75
h = 129.9 m
Hence, the height of kite above the ground = 129.9 m.

Que-4: If the shadow of a tower is 30 metres when the sun’s elevation is 30°, what is the length of a shadow when the sun’s elevation is 60° ?

Sol:  Let the height of the tower be x and shadow be y
In ΔABC
AB/BC = tan 30°
x/30 = tan 30° = 1/√3
x = 10√3
In ΔABD
AB/Ad = tan 60°
(10√3)/y  = tan 60° = √3
y = 10 m.

Que-5: The angle of elevation of the top of a tower (which is yet incomplete) at a point 120m from its base is 45°. How much higher should it be raised so that the elevation at the same point may become 60° ?

Sol:  Let PA be the unfinished tower.
Let B be the point of observation i.e. 120 m away from the base of the tower.
Now, AB = 120m
Let, ∠ABP = 45°
Let h m be the height by which the unfinished tower be raised such that its angle of
elevation of the top from the same point becomes 60°.
Let CA = h & ∠ABC = 60°
In triangle ABP,
tan45° = PA/AB
⇒ 1 = PA/120
⇒ PA = 120M
Now, in triangle ABC,
tan60° = CA/AB
√3 = (120+h)/120
h+120 = 120√3
h = 120(√3−1)
h = 120(1.732−1)  → (as√3=1.732)
h = 120×0.732
h = 87.84m

Que-6: Find the angle of depression from the top of a tower 140 m high of an object on the ground at a distance of 200 m from the foot of the tower.

Sol:  In triangle ABC,
tanθ = AB/BC
⇒ tanθ = 140/200 = 7/10 = 0.7
We have : tan35° = 0.7
Thus , the angle of depression is θ = 35°.

Que-7: A vertical tower is 20m high. A man at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower ?

Sol:  cos θ = 0.53
A man is standing at C
Let CB = x and height of tower AB = 20 m
In right ΔABC; we have
tanθ = AB/CB = 20/x   …(i)
cos θ = 0.53
From the table of cosines, we find that
θ = 58° (approx.)
Now substituting the value of θ in (i)
tan58° = 20/x
⇒ 1.6 = 20/x
∴ x = 20/1.6
= (20×10)/16
= 25/2
= 12.5 m

Que-8: In fig. 17.19, it is given that AB is perp. to BD and is of length x metres. DC = 30m, ∠ADB = 30° and ∠ACB = 45°. Without using tables, find x.

Sol:  In ΔABC, we have
tan45° = AB/CB = X/CB  …[∵tanθ = Perpendicular/Base]
1 = X/CB
⇒ CB = X   …(i)
Again in ΔADB, we have
tan30° = AB/DB = X/DB
⇒ 1/√3 = X/DB
DB = √3×X  …(ii)
But DB = DC + CB
∴ √3X = 30+X
⇒ √3X-X = 30
⇒ X(√3-1) = 30
⇒ X = 30/(√3-1) m
∴ X = {30(√3+1)}/{(√3-1)(√3+1)}
= {30(√3+1)}/(3-1) m
= {30(√3+1)}/2
= 15 (1.732 + 1) m.
= 15 × 2.732
= 40.98 m.

Que-9: In fig. 17.20, from a boat 200m away from a vertical cliff the angles of elevation of the top and the foot of a vertical concrete pillar at the edge of a cliff are 21° and  18°30′. Calculate :
(i) the height of the cliff   (ii) the height of the pillar

Sol:  Distance of boat from a vertical concrete pillar = 200m
Angle of elevation of the top and the foot = 21° and 18°30′
(i) Let the height of the cliff = h
tan 18°30′ = h / 200
0.33 = h / 200
h = 0.33 × 200
h = 66m
The height of the cliff = 66m

(ii) Let the height of the pillar = x
tan 21° = (x + 66) / 200
0.38 = (x + 66) / 200
x + 66 = 0.38 × 200
x + 66 = 76
x = 10m
The height of the pillar = 10m

Que-10: In Fig.17.21, the top of a tower is observed from two points on the same horizontal line through a point on the base of the tower. If the angles of elevation at the two points be 30° and 45°, and the distance between them is 20m, find the height of the tower.

Sol:  Let OB = x
∴ tan45° = h/x = 1 ; h = x
∴ CO = h+20
∴ tan30° = h/(h+20)
∴ 1/√3 = h/(h+20)
∴ √3h = h+20
∴ h = (20(√3−1))m
∴ h = 10(√3+1)m
h = 27.32 m

Que-11: 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°. When he moves 40m away from the bank, he finds the angle of elevation to be 30°. Find :
(i) the height of the tree, correct to 2 decimal places   (ii) the width of the river

Sol:  Let AB be the tree of height ‘h’ m and BC be the width of the river.
Let D be the point on the opposite bank of tree such that CD = 40 m.
Here ∠ADB = 30° and ∠ACB = 60°
Let speed of the boat be ‘x’ metre per minute.
In ΔABC,
AB/BC = tan60° = √3
⇒ h/BC = √3
⇒ h = BC√3
In ΔADB,
AB/BD = tan30°
⇒ h/(40+BC) = 1/√3
⇒ BC√3/(40+BC) = 1√3
⇒ BC√3⋅3 = 40+BC
⇒ 3BC = 40 + BC
⇒ 3BC – BC = 40
⇒ 2BC = 40 m
⇒ BC = 40/2 m
⇒ BC = 20 m
∴ h = 20 × 1.732 = 34.64 m
Hence, height of the tree is 34.64 m and width of the river is 20 m.

Que-12: In Fig. 17.22, the angle of elevation of the top P of a vertical tower PQ from a point X is 60°; at a point Y, 40m vertically above X, the angle of elevation is 45°.
(i) Find the height of the tower PQ   (ii) Find the distance XQ

Sol:  We have
XY = 40m, ∠PXQ = 60° and ∠MYQ = 45°
Let PQ = h
Also, MP = XY = 40m, MQ =  PQ  – MP = h – 40
(i) In ΔMYQ,
tan45° = MQ/MY
⇒ 1 = (h-40)/MY
⇒ MY = H – 40
⇒ PX = MY = h – 40                       …………….(1)
Now , in ΔMXQ,
tan60° = PQ/PX
⇒ √3 = h/(h-40)                         [From (i)]
⇒ h√3 – 40√3 = h
⇒ h√3 – h = 40√3
⇒ h (√3-1) = 40√3
⇒ h = 40√3/(√3-1)
⇒ h = {(40√3)/(3-1)} × {(√3+1)/(√3+1)}
⇒ h = {40√3(3+1)}/(3-1)
⇒ h = {40√3(√3+1)}/2
⇒ h = 20√3(3+1)
⇒ h = 60+20√3
⇒ h = 60 + 20×1.73
⇒h = 60 + 34.6
∴ h = 94.6m = 95 m
So, the height of the tower PQ is 95 m.

(ii)In ΔYRQ, we have
tan 45° = QR/YR
1 = x/YR
YR = x or XP = x [because YR=XP] —- (1)
Now In ΔXPQ we have
tan60° = PQ/PX
√3 = (x+40)/x (from equation 1)
x(√3−1) = 40
x = 40/(√3−1)
x = 40/(1.73−1) = 54.79 m
Distance XQ = 55 m.

Que-13: A guard observe an enemy boat, from an observation tower at a height of 180m above sea-level, to be at an angle of depression of 29°.
(i) Calculate, to the nearest metre, the distance of the boat from the foot of the observation tower.   (ii) After some time, it is observed that the boat is 200m from the foot of the observation tower. Calculate the new angle of depression.

Sol: (i)  Tan 29° = Height of Tower / Distance of boat from foot
=> 0.5543  = 180/Distance of boat from foot
=> Distance of boat from foot = 324.73 m
distance of the boat from the foot of the observation tower  = 325 m

(ii) Tan α   = Height of Tower / Distance of boat from foot
=> Tan α   = 180/ 200
=> Tan α   = 0.9
=> α   = Tan⁻¹(0.9)
=> α   = 41.99°
the new angle of depression = 42°

Que-14: In Fig. 17.23, the shadow of a vertical tower on level ground increase by 10m, when the altitude of the sun changes from 45° to 30°. Using the given figure, find the height of the tower, correct to one place of decimals.

Sol:  Let the height of tower be h meter and length of shadow y meter initially.
In ΔABC,
tan 45° = AB/BC
1 = h/y
y = h     …(1)
In ΔABD,
tan 30° = AB/DB
1/√3 = h/(y+10)
y + 10 = h√3    …(2)
Put y = h in equation (ii),
h + 10 = h√3
h(√3-1) = 10
h = {10(√3+1)}/{(√3-1)/(√3+1)}
h = 103-1(3+1)
h = 102(3+1)
h = 5(1.732 + 1)
h = 5 × 2.732
h = 13.66 = 13.7 m.

Que-15: A player sitting on the top of a tower of height 20m observes the angle of depression of a ball lying on the ground as 60°. Find the distance between the foot of the tower and the ball.

Sol:  Let C be the point where the ball is.
C = 60o (alternate angles)
In ABC, tan 60o = AB/BC
√3 = 20/x
x = 20/√3
= 20 (√3/3)
= 11.55 m

Que-16: The angle of elevation of the top of a tower from two points at distances a and b metres from the base and in the same straight line with it are complementary. Prove that the height of the tower is √ab metres.

Sol:  Let AB be the tower and let C and D be the two points of the observer. Then,
AC = a metres and AD = b metres
Let ∠ACB = θ. Then, ∠ADB = (90°−θ).
Let AB = h metres
In right △CAB, we have
tanθ = AB/AD
tanθ = h/a
h = a tanθ ……(1)
In right △DAB, we have,
tan(90°−θ) = AB/AD
cotθ = h/b
h = b cotθ …….(2)
From 1 and 2, we get,
h² = ab
h = √ab metres

Que-17: The length of a string between a kite and a point on the ground is 90 metres. If the string makes an angle θ with the level ground such that tan θ = 15/8, how high is the kite? Assume, there is no slack in the string.

Sol:  Length of string between the point on ground and kite =  90
Angle made by the string with the ground is θ and tanθ=158
⇒ θ = tan¯¹(15/8)
The height of the kite be H m
If we represent the above data in the figure as shown then it forms right-angled triangle
ABC.
We have,
tanθ = AB/BC
⇒ 15/8 = H/BC
⇒BC =  84/15 …….(1)
In ΔABC ,  by Pythagoras theorem we have
AC² = BC²+AB²
⇒ 90² = (8H/15)²+H²
⇒ 90² = {(8H)²+(15H)²}/15²
⇒ {H²(8²+15²)} = 90²×15²
⇒ {H²(64+225)} = (90×15)²
⇒ H² = {(90×15)²}/289
= H² = {((90×15)/17)}²
⇒ H = (90×15)/17 = 79.41
∴ height of kite from ground H = 79.4 m

Que-18: An observer in Fig. 17.25, {1*(1/2)}m tall, is 28.5 m away from a tower 30m high. Determine the angle of elevation from his eyes to the top of the tower.

Sol: Here, let DC be an observer and θ is the angle of elevation of the top of the tower from his eye.
⇒ AB is height of tower and CB is distance between tower and observer.
⇒ CB = DE = 28.5m
⇒ So in △ADE, tanθ = AE/DE
⇒ tanθ = (30−1.5)/28.5
⇒ tanθ = 28.5/28.5
⇒ tanθ = 1
∴ θ = tan¯¹(1)
∴ θ = 45°

Que-19: Two men are on diametrically opposite side of a tower. The measures the angle of elevation of the top of the tower as 20° and 24° respectively. If the height of the tower is 40m, find the distance between them. 

Sol:  In △ABD,
tan20° = AD/BD
⇒ 0.364 = 40/BD (∵tan20°=0.364)
⇒ BD = 40/0.364
⇒ BD = 109.89
In △ACD
tan24° = AD/CD
⇒ 0.445 = 40/CD (∵tan24°=0.445)
⇒ CD = 40/0.445
⇒CD = 89.89
Therefore,
Required distance between the two men = BD+CD = (109.89+89.89) m = 199.78m
Hence the distance between the two men is 199.78m.

Que-20: The horizontal distance between the two towers is 60m and the angular depression of the top of the first, as seen from the top of the second, which is 150m high is 30°; find the height of the first.

Sol:  As the Angle of Depression is given from the top of Second Tower
Hence, Second Tower is Larger than the First,so we can clearly see that, Height of First Tower must be less than the Height of the Second Tower (=150 m)
We can see that, BCDE is a Rectangle
so, BC = ED = 60
BE = CD = h (Let)
Now, in △AED
tan30° = AE/ED
⇒ 1/√3 = AE/60
⇒ AE = 60/√3
So, Height of First Tower is h = BE = AB−AE
= 150−(60/√3)
= 150-34.64 = 115.36 m

Que-21: From the top of a cliff, 150 metre high, the angle of depression of the top and bottom of a tower are observed to be 30° and 60°, find the height of the tower.

Sol: BF = (150−x)m. Let AC=y
tan30° = (150−x)/y
tan60° = 150/y
⇒ y = (150−x)√3 = 150/√3
∴ 150−x = 50 or x = 100 m

Que-22: From a point P on a level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100m high, how far is P from the foot of the tower ?

Sol:  Let height of tower AB is 100 m.
The distance between P from foot of tower is PB.
In ΔPAB
tan 30 = AB/PB
⇒ 1/√3 = 100/PB
⇒ PB = 100√3
⇒ PB = 100 × 1.732
⇒ PB = 173.2

Que-23: From the top of a building AB, 60 metres high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find
(i) the horizontal distance between AB and CD;
(ii) the height of the lamp post CD

Sol:  Let AE = h, AC = x, and AC = ED.
it is also given AB = 60 m. Then BE = 60 – h And
∠ACB = 60°, ∠BDE = 30°
(i) So we use trigonometric ratios.
In Δ ABC
⇒ tan60° = AB/AC
⇒ √3 = 60/x
⇒ x = 60/√3
⇒ x = 34.64
Hence the distance between AB and CD is 34.64

(ii) Again in Δ BDE
⇒ tan30° = BE/DE
⇒1/√3 = (60-h)/x
⇒ 60/√3 = (60-h)√3
⇒ 60 = 180-3h
⇒3h = 180-60
⇒3h = 120
⇒h = 40
Hence the height of lamp post is 40 m

Que-24: An aeroplane when 3000m high passes vertically above another aeroplane at an instance when there angles of elevation at the same observation point are 60° and 45°. How many metres higher is the one than the other ?

Sol:  Let P1 and P2 denote the positions of the two planes. Then in right-angled ΔP1AB,
P1B/AB = tan45° ⇒ P1B = AB
In right-angled ΔP2AB,
⇒ P2B/AB = tan60° = 3
⇒ AB = P2B/√3
⇒ AB = 3000/√3
⇒ AB = 1000√3
∴ Vertical distance between the two planes is P1P2 = P2B – P1B
= 3000 – 1000√3
= 1000( 3 – √3) m
= 1000 (1.268) = 1268 m.

Que-25: The angle of elevation of a cloud from a point 60m above the lake is 30° and the angle of depression of the reflection of the cloud in the lake is 60°. Find the height of the cloud.

Sol:  Let C be the cloud and D be its reflection . Let the height of the cloud is h metres .
BC = BD = h
BQ = AP = 60m
∴ CQ = h – 60 and DQ = h + 60
In ΔCQP,
PQ/CQ = cot30°
⇒ PQ/(h-60) = √3
⇒ PQ = √3(h-60)   ….(i)
In ΔDQP,
PQ/DQ = cot60°
⇒ PQ/(h+60) = 1/√3
⇒ PQ = 1/√3(h+60)   ..(ii)
From (i) and (ii),
⇒ √3(h-60) = 1/√3(h+60)
⇒ 3h – 180 = h + 60
⇒ 2h = 240
⇒ h = 120.

Que-26: The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If the tower is 50m high, what is the height of the hill?

Sol:  In ΔABD
tan30° = AB/BD
⇒ 1/√3 = 50/BD
⇒BD = 50√3
In ΔCDB
tan60° = CD/BD
⇒ √3 = CD/50√3
=> CD = 150
Therefore, the height of the hill is 150 m

–: End of Heights and Distances Class 10 OP Malhotra Exe-17 ICSE Maths ICSE Maths Solutions :–

Return to :  OP Malhotra S Chand Solutions for ICSE Class-10 Maths

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