Height and Distance Class 10 RS Aggrwal ICSE Maths Solutions ch-23. In this article you will get practice questions / problems on Height and Distance with solutions / answer. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.

Height and Distance Class 10 RS Aggrwal ICSE Maths Solutions ch-23

Board	ICSE
Publications	Goyal Brothers Prakashan
Subject	Maths
Class	10th
Chapter-23	Height and Distances
Writer	RS Aggarwal
Book Name	Foundation
Topics	solving questions on height and distance

Practice Questions on Height and Distance with Solutions

Class 10 RS Aggrwal ICSE Maths Solutions ch-23

Que-1: The angle of elevation of the top of a pole from a point on the level ground and 15 cm away from the pole ie 30º . Find the height of the pole .

Sol: let AB = tower
CD = building
such that ∠ACD = 30º
∠ADB = 60º
AE = h m (say)
EB = CD = 15 m
and BD = x m (say)
= CE
Now , in ΔAEC , ∠E = 90º we have ,
tan30º = h/x
1/√3 = h/x
x = h√3 … (i)
Again in ΔABD , ∠E = 90º we have
tan 60º = AB/BD
√3 = h+15/x
√3(√3.h) = h+15 …[from (i) x = √3.h]
⇒ 3h − h = 15
⇒ h = 7.5 m
⇒ x = √3⁢ℎ =√3 ×7.5 ≈1.732 ×7.5 m
x ≈ 12.99 m
Hence height of the tower = 7.5 + 15 = 22.5 m
and distance between tower and building = 12.99 m

Que-2: From the top of a a cliff , 50 m high , the angle of depression of a buoy is 30º . Calculate to the nearest metre , the distance of the buoy from the foot of the cliff [Take √3 = 1.732] .

Sol: Angle of depression = angle of elevation = 30º
Let the horizontal distance from the foot of the cliff to the buoy be 𝑥
tan 30º = opposite/adjacent = 50/𝑥
Given:
tan 30º  = 1/√3 = 1/1.732 ≈ 0.577
So,
0.577 = 50/𝑥
​x = 0.577/50 ≈ 86.65

Distance ≈ 87 metres

Que-3: A vertical pole is 12 m high and the length of it’s shadow is 12√3 m. What is the angle of elevation of the sun ?

Sol: Let the angle of elevation of the sun be θ.
Pole height = 12 m
Shadow length = 12√3
Using the tangent ratio:
tan 𝜃 = opposite/adjacent = 12/12√3 = 1/√3
We know:
tan 30º = 1/√3
So,
𝜃 = 30º

Que-4: A kite is flying with a thread 80 m long . If the thread is assumed stretched straight and makes an angle of 60º with the horizontal , find the height if the kite above the ground ?

Sol: Angle of elevation = 60º
Length of thread = 80 m (hypotenuse)
Height of kite = opposite side
sin 60∘ = height/80
√3/2 = ℎ/80
ℎ = 40√3 ≈ 69 m
Height of the kite = 69 m (approx.)

Que-5: The length of a string between a kite and a point on the ground is 85 m. If the string makes an angle θ with the level ground such that tanθ = 15/8 , how high is the kite ?

Sol: Given:
tanθ = 15/8
String length (hypotenuse) = 85 m
Let height of kite = opposite side.

Form a right triangle:
For tanθ = 15/8:
Opposite side = 15k
Adjacent side = 8k
Hypotenuse = √(15²+8²) k = √(289) 𝑘 = 17𝑘
Given hypotenuse = 85 m:
17𝑘 = 85 ⇒ 𝑘 = 5
Height (opposite side):
Height=15k=15×5=75 m

Que-6: A vertical tower is 20 m high . A  man standing 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
tan⁡𝜃 =𝐴⁢𝐵/𝐶⁢𝐵 =20/𝑥   …(i)
cos θ = 0.53
From the table of cosines, we find that
θ = 58º (approx.)
Now substituting the value of θ in (i)
tan⁡58º=⁢20/𝑥
⇒1.6 =20/𝑥
∴ 𝑥 =20/1.6
= (20×10)/16
= 25/2
= 12.5 m

Que-7: At a point on a level ground , the angle of elevation of the top of a tower is θ such that tan θ = 7/12 . On walking 64 m towards the tower , the angle of elevation is Φ , where tan Φ = 3/4 . Find the height of the tower.

Sol: Let the height of the tower be h and the initial distance from the tower be x.
Given:
tanθ = 7/12 = h/x ⇒ h = (7/12)x …(1)

After walking 64 m towards the tower, the distance becomes x−64.

tanΦ = 3/4 = ℎ/(𝑥−64) ⇒ ℎ = (3/4)(x-64)…(2)

Equate (1) and (2):
(7/12)x = 3/4(x-64)

Multiply both sides by 12:
7𝑥 = 9(x-64)
7x = 9x−576
576 = 2x
x = 288 m

Find height:

Using ℎ = (7/12)𝑥
h = 7/12 × 288
h = 7 × 24 = 168 m

Que-8: From the two points A and B on the same side ofd a building , the angles of elevation of the top of the buildings are 30º and 60º respectively . If the height of the building is 10m , find the distance between A and B , correct to two decimals places.

Sol: Let CD is building A and B are two given points using horizontally on the same side of building.
In Δ DBC,
tan 60° = DC/CB
√3 = 10/𝑦 …..(1)
In Δ DCA,
tan 30° = DC/CA
1/√3 =10/𝑥+𝑦 …..(2)
From (1), put y = 10/√3 in (2), we get
1/√3 =10/(𝑥+(10/√3))
1/√3 =10⁢√3/(√3⁢𝑥+10)
30 = √3x + 10
x = 20/√3
x = 11.55 m.

Hence, distance between two points A and B is 11.55 m.

Que-9: The shadow of a verticval tower AB on level ground is increased by 10 m , when the altitude of the sun changes from 45º to 30º , as shown in the figure . FInd the height of the tower and give your answer correct to 1/10 metre.

Sol: Let the height of tower be h meter and length of shawdow 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 = (10/(3−1⁢))(√3+1)
h = (10/2⁢)(√3+1)
h = 5(1.732 + 1)
h = 5 × 2.732
h = 13.66 meter

Que-10: The angles of elevation of the top of a tower frokm two points on the ground at distances a metres and b metres from the base of the tower and in the same straight line with it are complementry . Prove that the height of the tower is √(ab) metres.

Sol: Let AB be the tower of height h metre
Let C and D be two points on the level ground such that BC = b metres, BD = a metres, ∠ACB = α and ∠ADB = β
Given, α + β = 90°
In ΔABC,
𝐴⁢𝐵/𝐵⁢𝐶 =tan⁡𝛼
⇒ℎ/𝑏 =tan⁡𝛼   …(i)
In ΔABD,
𝐴⁢𝐵/𝐵⁢𝐷 =tan⁡𝛽
⇒ℎ/𝑎 =tan⁡(90°−𝛼) =cot⁡𝛼  …(ii)
Multiplying (i) by (ii), we get,
(ℎ/𝑎)⁢(ℎ/𝑏) =1
⇒ h² = ab
∴ ℎ =√𝑎⁢𝑏 metre

Hence, height of the tower is √𝑎⁢𝑏 metre.

Que-11: A man in a boat rowing away from a lighthouse , 150 m high , takes 1.5 minutes to change the angle of elevation of the top of the lighthouse from 60º to 45º , Find the speed of the boat.

Sol: Let AB be the lighthouse and C and D be the two positions of the boat such that AB = 150 m, ∠ADB = 45° and ∠ACB = 60°.
Let speed of the boat be x metre per minute.
Therefore, CD = 2x m;
In ΔADB,
𝐴⁢𝐵/𝐷⁢𝐵 =tan⁡45º
⇒ BD = 150 m

In ΔABC,
𝐴⁢𝐵/𝐵⁢𝐶 =tan⁡60º=⁢√3
 ⇒ 150/𝐵⁢𝐶 =√3
 ⇒ 𝐵⁢𝐶 =150/√3
= 150/1.732
= 86.605 m
∴ CD = BD – BC
= 150 – 86.605
= 63.395 m
⇒ 2x = 63.395
 ⇒𝑥 =63.395/2
= 31.6975 m/min
= 31.6975/60 m/sec
= 0.53 m/sec

Hence, the speed of the boat is 0.53 m/sec  

Que-12: Two pillars of equal heights stand on either side of a roadway , which is 120 m wide . At a pint on the road lying between the pillars , the elevations of the pillars are 60º and 30º respectively. Find the height of the pillar and the position of the point .

Sol: Let AB and CD be the two towers of height h m.
Let P be a point in the roadway BD such that BD = 150 m, ∠APB = 60° and ∠CPD = 30°
In ΔABP,
𝐴⁢𝐵/𝐵⁢𝑃 =tan⁡60º
⇒𝐵⁢𝑃 =ℎ/tan⁡60º =ℎ/√3

In ΔCDP,
𝐶⁢𝐷/𝐷⁢𝑃 =tan⁡30º
⇒𝑃⁢𝐷 =1/√3

Now, 120 = BP + PD
⇒120 =ℎ/√3 +1/√3
∴ ℎ = 120/√(3+(1/√3))
= 120/2.309
= 51.97 m
Hence, height of the pillars is 51.97 m.
The point is 𝐵⁢𝑃/√3 from the first pillar.
That is the position of the point is 51.97/√3⁢𝑚 from the first pillar.
The position of the point is 30.005 m from the first pillar.

Que-13: The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60º. At a point Y, 40 m vertically above X , the angle of elevation is 45º . Find the height of tower PQ and 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
In ΔMYQ,
tan⁡45 ° =𝑀⁢𝑄/𝑀⁢𝑌
⇒1 = (ℎ−40)/𝑀⁢𝑌
⇒ MY = H – 40
⇒ PX = MY = h – 40                       …………….(1)

Now , in ΔMXQ,
tan⁡60 ° =𝑃⁢𝑄/𝑃⁢𝑋
⇒√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

So, the height of the tower PQ is 94. 6 m.

Que-14: A man 1.8 m tall stands at a distance of 3.6 m from a lamp post and casts a shadow of 5.4 m on the ground . Find the height of the lamp post.

Sol: Let AB be the lamp post whose height be (h) meter as shown in figure.
Let DC be the man standing at 3.6 meter from the lamp post whose height is 1.8 meter as shown in figure.
The CE is the shadow of the man whose length is 5.4 meter as shown in figure.

let ∠CED = xº
tan xº = CD/CE = 1.8/5.4 = 1/3

Now in ∠ABE ,
tan xº = AB/BE = h/BC+CE

So,
1/3 = h/3.6+5.4
h = 9/3
h = 3 m

∴ Height of the lamp post is 3 m.

Que-15: In the adjoining figure , a man stands on the ground at a point A , which is on the same horizontal plane as B , the foot of the vertical pole BC . The height of the pole is 10 m . The man’s eye is 2 m above the ground . He observes the angle of elevation of C , the top of the pole as xº , where tanxº = 2/5 . Calculate the distance AB in metres.

Sol: Let AD be the height of the man, AD = 2 m
∴ CE = (10 – 2) = 8 m

In ΔCED,
𝐶⁢𝐸/𝐷⁢𝐸 =tan⁡𝑥∘=⁢25
⇒8𝐷⁢𝐸 =25
⇒ DE = 20 m
Here AB = DE

∴ AB = 20 m

∴ The distance AB is 20 m.

Que-16: From a window A , 10 m above the ground , the angle of elevation of the top C of a tower is xº , where tanxº = 5/2 and the angle of depression of the foot D of the tower is yº , where tan y = 1/4 .

Sol: Let CD be the height of the tower
And height of window A from the ground = 10 m
In right ΔAEC, we have
tan⁡𝑥∘=⁢𝐶⁢𝐸/𝐴⁢𝐸
⇒5/2 =𝐶⁢𝐸𝐴/⁢𝐸   …(∵tan⁡𝑥=5/2)
∴ 𝐴⁢𝐸 =2⁢𝐶⁢𝐸/5  …(i)

In right ΔAED, we have
tan⁡𝑦∘=⁢𝐸⁢𝐷/𝐴⁢𝐸
⇒1/4 =10/𝐴⁢𝐸
∴ AE = 40

Now substituting the value AE in (i), we get
(2/5⁢)𝐶⁢𝐸 = 40
⇒𝐶⁢𝐸 =40×5/2 =100 𝑚
∴ Required height of tower
CD = CE + ED
= (100 + 10) m
= 110 m

Que-17: An aeroplane at an altitude of 900 m finds that two ships are sailing towards it in the same direction . The angle of depression of the ships , as observed from the plane are 60º and 30º respectively . Find the distance between the ships.

Sol: C and D are the positions of the two ships.
Let the distance between the two ships be “x”
Let BC = y
∴ BD = (x + y)

In the right ∆ABC, tan 30° = AB/BD
1/√3 =900/𝑥+𝑦
x + y = 900⁢√3
y = 900⁢√3 −𝑥  …(1)

In the right ∆ABC, tan 60° = AB/BC
√3 = 900/𝑦
y = 900/√3  …(2)
From (1) and (2) we get
900/√3 = 900√3 −𝑥
900 = 900 ×3 −√3⁢𝑥
√3⁢𝑥 = 2700 – 900
x = 1800/√3
= 1800×√3/√3×√3
= (1800×√3)/3
= 600 × 1.732
= 1039.2 m

Distance between the two ships = 1039.2 m

Que-18: In the given figure , AB is a tower and two objects C and D are located on the ground on the same side of AB . When observed from the top B of the tower , their angles of depression are 45º and 60º respectively . Find the distance between the objects , if the height of the tower is 180 m .

Sol: Let CD = x, BD = y, ∠ADB = 60°, ∠ACB = 45°
So we use trigonometric ratios.

In a triangle ABC,
tan⁡45° =⁢ 180/𝑥+𝑦
⇒𝑥 + 𝑦 = 180   ……(1)

Again in a triangle ABD,
tan⁡60° =⁢ 180/𝑦
⇒√3 =180/𝑦
⇒ y =180/√3  ……(2)

So from (1) and (2) we get
𝑥 +180/√3 = 180
⇒√3⁢𝑥 =180(√3−1)
⇒𝑥 = 76.07

Hence the required distance is approximately 76.07 m

Que-19: From the top of a church spire 96 m high , the angles of depression of two vehicles on a road , at the same level as the base of the spire and on the same side of it are xº and yº, where tan xº = 1/4 and tan yº = 1/7 . Calculate the distance between the vehicles .

Sol: Thus, the distance between the cars will be equal to the difference in the distances between the church bottom and the first car and the church bottom and the second car which will be equal to (y-x) m
tanxº = 1/4
tanyº = 1/7

In ΔABD ,
tanxº = 96/x
96/x = 1/4
x = 384 m

Now in  ΔABC ,
tanyº = 96/y
1/7 = 96/y
y = 672 m

y-x = 672 – 384 = 288 m

∴ The distance between the cars is 288 m.

Que-20: Two men standing on the same side of a tower in a straight line with it , measure the angles of elevation of the top the tower as 25º and 50º respectively . If the height of the tower is 70 m , find the distance between the two men.

Sol: Let CD be the distance between the two persons.
In ΔABC,
cot 50° = BC/AB
cot (90° – 40°) = BC/70
BC = 70 tan 40°
BC = 70 x 0.8391 = 58.74 m

In ΔABD,
cot 25° = BD/AB
cot (90° – 65°) = BD/70
tan 65° = BD/70
BD = 70 tan 65°
BD = 70 x 2.11451 = 150.12 m
CD = 150.12 – 58.74 = 91.38 m

∴ The distance between the two person be 91.38 m.

Que-21: In the figure , AB represents a pole and CD represents a 60 m high tower , both of which are standing on the same horizontal plane . From the top of the tower , the angles of depression of the top and the foot of the pole are 30 and 60 respectively. Calculate : (i) the horizontal distance between the pole and the tower . (ii) the height of the pole

Sol: Given that AB is a building that is 60 m, high.
Let BC = DE = x and CD = BE = y
⇒ AE = AB – BE = 60 – y

i. In right ΔAED,
tan⁡30∘=⁢𝐴⁢𝐸/𝐷⁢𝐸
⇒1/√3 =60−𝑦/𝑥
⇒𝑥 =60⁢√3 −𝑦⁢√3   …(1)

In right ΔABC,
⇒tan⁡60∘=⁢𝐴⁢𝐵/𝐵⁢𝐶
⇒√3 =60/𝑥
⇒𝑥 =60/√3
⇒𝑥 =60/√3 ×√3/√3
⇒𝑥 =60⁢√3/3
⇒𝑥 =20⁢√3
⇒ x = 20 × 1.732
⇒ x = 34.64 m

Thus, the horizontal distance between AB and CD is 34.64 m.

ii. From (1), we get the height of the lamp post = CD = y
𝑥 =60⁢√3 −𝑦⁢√3
⇒20⁢√3 = 60⁢√3 −𝑦⁢√3
⇒ 20 = 60 – y
⇒ y = 60 – 20
⇒ y = 40 m

Thus, the height of the lamp post is 40 m.

Que-22: From a boat 200 m away from a vertical cliff , the angles of elevation of the top and the foot of a vertical pillar at the edge of the cliff are 36 and 34 respectively . Find : (i) the height of the cliff and (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

Therefore, The height of the cliff = 66m and The height of the pillar = 10m

Que-23: A man on the top of a vertical observation tower , observers a car moving at a speed coming directly towards it . If it takes 10 minutes for the angle of depression to change from 30 to 45 , how soon after this , will the car reach the observation tower ?

Sol: Here, ∠ACB = 30° and ∠ADB = 45°.
Let C denote the initial position of the car and D be its position after 12 minutes.
Let the speed of the car be x meter/minute, then
CD = 12x meters     …..( ∵ Distance = speed x Time)
Let the car take t minutes to reach the tower from D.
Then, DB = tx meters

Now in the right-angled triangles ACB,
tan 30° = AB/BD
⇒ 1/√3 =AB/(BC+CD)
⇒ 1√3 =AB/(12⁢𝑥+𝑡⁢𝑥)
⇒ AB = (12⁢𝑥+𝑡⁢𝑥)/√3         ….(1)

Also, in the right-angled triangle ADB,
tan 45° = AB/DB
⇒ 1 = AB/DB
⇒ AB = DB = tx    ……(2)

From (1) and (2), we have
t = 12/√3−1 =12(⁢√3+1)/2
t = 6⁢(√3+1)
t = 15.39

∴ Time = 16.39 minutes
Time = 16 minutes 23 seconds.

Que-24: The angle of depression of a stationary cloud from a point 35 m above a lake is 30 and the angle of depression of it’s reflection in the lake is  60 . What is the height of the cloud above the lake-level ?

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 = 35m
∴ CQ = h – 35 and DQ = h + 35

In ΔCQP,
PQ/CQ =cot⁡30∘
⇒ PQ/h−35 =√3
⇒ PQ =√3⁢(h−35)   ….(i)

In ΔDQP,
PQ/DQ =cot⁡60∘
⇒ PQ/h+35 =1/√3
⇒ PQ =1√3⁢(h+35)   ..(ii)

From (i) and (ii),
⇒ √3⁢(h−35) =1/√3⁢(h+35)
⇒ 3h – 105 = h + 35
⇒ 2h = 140
⇒ h = 70

Thus , the height of the cloud is 70 m.

Que-25: (i) A tower subtends an angle α on the same level as the foot of the tower and at a second point h metres above the first , the depression of the foot of the tower is β . Show that the height of the tower is h tan α cot β.
(ii) The angle of elevation from a point P of the top of a tower QR , 50 m high is 60 and that of the tower PT from a point Q is 30 . Find the height of the tower PT , correct to the nearest metre .

Sol: (i) In ΔABD ,
H = xtanα …(i)

In ΔBCE ,
x = hcotβ …(ii)

from (i) and (ii) ,
H = htanαcotβ

(ii) In ΔPQR,
⇒tan ⁡P = QR/PQ
⇒tan⁡ 60° =50/PQ
⇒√3 =50/PQ
⇒PQ = 50/√3

Again in  Δ PQT,
tan ⁡Q =PT/PQ
⇒tan⁡ 30º = PT/(50/√3)                  [Using (1)]
⇒1/√3 =PT/(50/√3)
⇒PT =50/3
PT = 16.66
PT = 17 m (approx)

Que-26: A man observes the angle of elevation of the top of the tower to be 45 . He walks towards it in a horizontal line through its base . On covering 20 m , the angle of elevation changes to 60 . Find the height of the tower correct to 2 significant figures.

Sol: Let the height of the tower be ‘h’ m.

In Δ ADC ,
tan 45° = ℎ/20+𝑥
1 = ℎ/20+𝑥
⇒ h = 20 + x

Also , In ΔBDC ,
tan 60° = ℎ/𝑥
√3 = ℎ/𝑥
⇒ x = ℎ/√3              …(2)
h = 20 + ℎ/√3
ℎ −ℎ/√3 =20
ℎ⁡((√3−1)/√3) =20
ℎ =20⁢√3/(√3−1) ×(√3+1)/(√3+1)
=20⁢(3+√3)/3−1
=20⁢(3+1.732)/2
= 10 (4.732)

Height of the tower . h = 47.32 m

Que-27: From the top of a cliff 60 m high , the angles of depression of two boats are 30 and 60 respectively . Find the distance between the boats , when the boats are : (i) on the same side of the cliff , (ii) on the opposite sides of the cliff.

Sol: Let the height of the cliff be h = 60 m.
Let the distances of the two boats from the base of the cliff be x1 and x2 respectively.

for first boat :
tan(30°)= x1/h
1/√3= x1/60
​x1 = 60/√3
x1 = 60/√3 × √3/√3
x1 = 60√3/3
x1= 20√3
​
for second boat :
tan(60°)= x2/h
√3 = x2/60
x2 = 60√3

if boats are on same side of the cliff distance between the boats = 20√3+60√3 = 80√3
if boats are on opposite side of the cliff distance between the boats = 60√3-20√3 = 40√3

Que-28: From the top of a hill the angles of depression of two consecutive kilometer stones , due cast are found to be 30 and 45  respectively . Find the distance of the two stones from the foot of the hill .

Sol: Let AB be the hill of height ‘h’ km and C and D be two consecutive stones such that CD = 1 km, ∠ACB = 30° and ∠ADB = 45°
In ΔABD,
𝐴⁢𝐵/𝐵⁢𝐷 =tan⁡45∘=⁢1
⇒ BD = h

In ΔABC,
𝐴⁢𝐵/𝐵⁢𝐶 =tan⁡30∘
⇒ℎ/𝐵⁢𝐶 =1/√3
⇒ ℎ/ℎ+1 =1/√3
⇒ℎ =1/(√3−1)
= (√3+1)/2
= 2.732/2
= 1.366 km

∴ BD = 1.366 km
BC = BD + DC
= 1.366 + 1
= 2.366 km

Hence, the two stone are at a distance of 1.366 km and 2.366 km from the foot of the hill.

Que-29: An aeroplane at an altitude of 1500 m finds that two ships are sailing towards it in the same direction . The angles of depression as observed from the aeroplane are 45 and 30 respectively . Find the distance between the two ships .

Sol: A is the aeroplane, D and C are the ships sailing towards A. Ships are sailing towards the aeroplane in the same direction.
In the figure, height AB=1500 m

In the right-angled ΔABC
tan⁡45º=⁢𝐴⁢𝐵/𝐵⁢𝐶
⇒1 = 1500/𝐵⁢𝐶
=> BC = 1500 m

In the right-angled ΔABD,
tan⁡30º =⁢ 𝐴⁢𝐵/𝐵⁢𝐷
1/√3 = 1500/BD
BD = 1500√3

=> BD = 1500(1.732)  = 2598 m
∴ Distance between the ships = CD = BD – BC= 2598 – 1500
= 1098 m

Que-30: An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45 and 60 respectively . Find the width of the river .

Sol: Let A be the position of the airplane and let BC be the river. Let D be the point in BC just below the airplane.
B and C be two boats on the opposite banks of the river with angles of depression 60° and 45° from A.

In ΔADC,
tan⁡45º=⁢𝐴⁢𝐷/𝐷⁢𝐶
⇒1 =250/𝑦
⇒ y = 250 m = DC

In ΔADB,
tan⁡60º=⁢𝐴⁢𝐷/𝐵⁢𝐷
⇒√3 =250/𝑥
⇒𝑥 =250/√3
= 250⁢√3/3
= 250×1.732/3
= 144.3 m = BD

∴ BC = BD + DC
= 144.3 + 250
= 394.3 ≈ 394 m

Thus, the width of the river is 394 m.

Que-31: From the top of a tower , 100 m high , a man observes the angles of depression of two ships A and B , on opposite sides of the tower are 45 and 38 respectively . If the foot of the tower and the ships are in the same horizontal line , find the distance between the two ships A and B to the nearest metre . 

Sol: Let AC = x,
BC = y
Now ∠A = 45°   …(Alternative angles)
and ∠B = 38°   …(Alternative angles)

In ΔACD,
tan 45° = 100𝑥
⇒ 1 = 100/𝑥
∴ x = 100   …(1)

In ΔBCD,
tan 38° = 100/𝑦
0.7812 = 100/𝑦
y = 100/0.7812 ≈128

Hence Distance between two ships A, B
= x + y
= 100 + 128
= 228 m

Que-32: The horizontal distance between two towers is 120 m . The angles of elevation of the top and angle of depression of the bottom of the first tower as observed from the second tower is 30 and 24 respectively . Find the heights of the two towers . Give your answers to 3 significant figures .

Sol: In ΔAEC
tan 30º = (AE)/(EC)
=> 1/√3 = (AE)/120
=> 120/√3 = AE
∴ AE = 69.28 m

In ΔBEC
tan 24º = (EB)/(EC)
=> 0.445 = (EB)/(120)
:. EB = 53.427 m`

Thus, height of first tower, AB = AE + EB
= 69.282 + 53.427
= 122.709 m
= 122 m (correct to 3 significant figures)

And height of second tower CD =  EB = 53.427 m = 53.4 (correct to 3 significant figures)

Que-33: From the top of a cliff , the angle of depression of the top and bottom of a tower are observed to be 45 and 60 respectively . If the height of the tower is 20 m , find : (i) the height of the cliff (ii) the distance between the cliff and the tower

Sol: From figure,
∠ACE = ∠FAC = 45° (Alternate angles are equal)
∠ADB = ∠FAD = 60° (Alternate angles are equal)

Let BD = x.
From figure,
EC = BD = x meters.
EB = CD = 20 meters.

In △ AEC,
⇒ tan 45° = EC/AE
​⇒ 1 = x/AE
​⇒ AE = x meters.

In △ ABD,
⇒tan 60°= AB/BD
⇒√3 = AE+EB/x
⇒ √3 = x+20/x
√3x = x+20
x(√3-1) = 20
x = 20/(√3-1)
x = 20/0.732
= 27.32 m

From figure,
Height of cliff (AB) = AE + EB
= x + 20
= 27.32 + 20
= 47.32 meters.

Hence, the height of cliff = 47.32 meters.

(ii) From figure,
Distance between cliff and tower (BD) = x meters = 27.32 meters.
Hence, distance between cliff and tower = 27.32 meters.

Que-34: Two lamp posts AB and CD , each of height 100 m are on either side of the road . P is a point on the road between the two lamp posts . The angle of elevation of the top of the lamp post from the point P are 60 and 40 . Find the distances PB and PD .

Sol: Let PB = x and PD = y
tan(60°)= 100/PD
tan(60°)= 100/y
​y = tan(60°)100
y = 100/√3
y = 100/1.732
y = 57.74 m
PD = 57.74 m

tan(40°) = 100/PB
tan(40°) = 100/x
x = 100/tan(40°)
x = 100/0.8391
x = 119.19 m
PB = 119.19 m

— : Height and Distance Class 10 RS Aggrwal ICSE Maths ch-23 practice questions :–

Return to :–  RS Aggarwal ICSE Class 10 Solutions Goyal Brothers
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