RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 14 Height and Distance Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 14 Height and Distance Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance Maths book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 14
Chapter NameHeight and Distance
Exercise14
CategoryRS Aggarwal Solutions

RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance

Question 1:
Let AB be the tower standing on a level ground and O be the position of the observer. Then OA = 20 m and ∠OAB = 90° and ∠AOB = 60°
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 1.1
Let AB = h meters
From the right ∆OAB, we have
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 1.2
Hence the height of the tower is 203–√m=34.64m

Question 2:
Let OB be the length of the string from the level of ground and O be the point of the observer, then, AB = 75m and ∠OAB = 90° and ∠AOB = 60°, let OB = l meters.
From the right ∆OAB, we have
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 2.1

Question 3:
Let AB be the man,
AB= 1.6m, CD is the tower
AE CD, DE = AB
Let CE = h
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance Q3
In ∆ACE,
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 3.1
Height of tower = DE + DC = (1.6 + 25.98)m = 27.58 m

Question 4:
Let AB be the tree bent at the point C so that part CB takes the position CD, then CD = CB
Let AC = x meters
Then, CD = CB = (10 – x) m
and ∠ADC = 60°
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 4.1
Therefore, tree bent at the height of 4.64m from the bottom.

Question 5:
Let AB be the lamp post and CD be the boy, let CE be the shadow of CD
Let, ∠AEB = θ
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 5.1
From right ∆ECD, we get
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 5.2
From right ∆EAB, we get
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 5.3
Hence, the height of the lamp post = 2.5 m

Question 6:
Let CD be the height of the building
Then, ∠CAB = 30°, ∠CBD = 45°,
∠ADC = 90° and AB = 30m
CD = h meters and BD = x meters
From right ∆CAD, we have
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 6.1
From right ∆BCD, we have
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 6.2
From (1) and (2), we get
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 6.3
Putting h = 40.98m in (2), we get x = 40.98 m
Hence, height of building = 40.98m and
Distance of its base from the point A
= AB = (30+x) m
= (30+40.98) m = 70.98 m

Question 7:
Let CD be the tower and BD be the ground
Then, ∠CBD = 30°, ∠CAD = 60°
∠BDC = 90°, AB = 20 m, CD = h metre and AD = x metre
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 7.1
From ∆BCD
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 7.2
From right ∆CAD, we have
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 7.3
Hence, the height of the tower = 17.32m and the distance of the tower from the point A = 30m.

Question 8:
Let AB and CD be the building and the tower respectively.
AB = 15 m, AE ⊥ CD
ED = AB = 15 m
Let EC = h m
And BD = AE = x m
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 8.1
In CAE,
∠CAE = 30°and ∠AEC = 90°
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 8.2
In CBD, ∠CBD = 60° and ∠CDB = 90°
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 8.3
Eliminating x from (1) and (2), we get
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 8.4
Height of tower = CE + ED = (h + 15) m
= (7.5 + 15) m = 22.5m
Hence, Height of the tower = 22.5 m and the distance between the tower and the building = 12.99 m

Question 9:
AB and CD are the two houses.
Window is at A.
In ∆ ABD, ∠B = 90°, AB = 15m
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 9.1
AE is drawn perpendicular to CD
Therefore, AE = BD = 15 m
Let CE = h m
In ∆ ACE,
∠CAE = 30°, ∠CEA = 90°
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 9.2
Height of opposite house = CE + ED
= (h + 15) m = (8.66 + 15) m = 23.66 m
Hence proved.

Question 10:
Let AB be the tower with height = h m
AC = flag staff = x m
PB = 30 m
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 10.1
In ∆ PBC,
∠CPB = 60° and ∠CBP = 90°
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 10.2
Putting value of h in (1), we get
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 10.3
Thus, height of tower = 30m and height of flag staff = 21.96 m

Question 11:
Let AB be the tower h metre high. CA is the flag staff 5 meter high.
Let PB = x meter
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 11.1
In ∆ PBC,
∠CPB = 60°, ∠PBC = 90°
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 11.2
In ∆ APB,
∠APB = 30° and ∠ABP = 90°
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 11.3
Putting value of x in (1), we get
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 11.4
Thus, height of tower = 2.5m

Question 12:
Let SP be the statue and PB be the pedestal. Angles of elevation of S and P are 60° and 45° respectively.
Further suppose AB = x m, PB = h m
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 12.1
In right ∆ ABS,
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 12.2
In right ∆ PAB,
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 12.3
Thus, height of the pedestal = 2m

Question 13:
Let AB be the tower and let the angle of elevation of its top at C be 30°. Let D be a point at a distance 150 m from C such that the angle of elevation of the top of tower at D is 60°.
Let h m be the height of the tower and AD = x m
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 13.1
In ∆ CAB, we have
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 13.2
Hence the height of tower is 129.9 m

Question 14:
Let AB be the tower and BC be flagpole, Let O be the point of observation.
Then, OA = 9 m, ∠AOB = 30° and ∠AOC = 60°
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 14.1
From right angled ∆ BOA
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 14.2
From right angled ∆ OAC
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 14.3
Thus RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance Q14
Hence, height of the tower= 5.196 m and the height of the flagpole = 10.392 m

Question 15:
Let AB be the hill and let CD be the pillar. Draw DE AB, then, ∠ACB = 60° and ∠EDB = 30° and AB = 200 m
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 15.1
Height of the pillar = CD = 133.33 m
Distance of the pillar from the hill = ED = 2003√×3√3√ = 115.33m

Question 16:
Let AB be the height of the window of house and CD be another house on the opposite side of the street AC
Then, AB = 60 m
Draw BE ⊥ CD and join BC
Then, ∠EBD = 60° and ∠ACB = ∠CBE = 45°
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 16.1
From right ∆ CAB, we have
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 16.2
From right ∆ BED, we have
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 16.3
Hence, the height of the opposite house is 60(1+3–√)

Question 17:
Let O and B the two positions of the jet plane and let A be the point of observation.
Let AX be the horizontal ground.
Draw OC ⊥ AX and BD ⊥ AX.
Then, ∠CAO = 60°, ∠DAB = 30° and OC = BD = 1500√3 m
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 17.1
From right ∆ OCA, we have
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 17.2
From right ∆ ADB, we have
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 17.3
Thus, the aeroplane covers 3000 m in 15 seconds
Hence the speed of the aeroplane is RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 17.4

Question 18:
Let AB be the building and CD be the light house.
AE is drawn perpendicular to CD.
Now AB = 60 m
∠ADB = 60°, ∠CAE = 30°
Let BD = x m
AE = BD = x m
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 18.1
In right ∆ ACE, let CE = h
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 18.2
From (1) and (2),
203–√=3–√h
h = 20 m
Hence,
(i) Difference of heights of light house and building = 20m
(ii) The distance between light house and building = 34.64m

Question 19:
Let AB be the light house and let C and D be the positions of the ship.
Llet AD =x, CD = y
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 19.1
In ∆ BDA,
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 19.2
The distance travelled by the ship during the period of observation = 115.46 m

Question 20:

Let CD be the height of the building
Then, ∠CAB = 30°, ∠CBD = 45°, ∠ADC = 90° and AB = 30m
CD = h metres and BD = x metres.
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 20.1
From right ∆ CAD, we have
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 20.2
From right ∆ BCD, we have
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 20.3
Putting h = 40.98 in (2), we get x = 40.98 m
Hence height of building = 40.98 m and Distance of its base from the point
A = AB = (30 + x) m
= (30 + 40.98) m = 70.98 m

Question 21:
Let CD be a tree. Angle of elevation from A and B are 60° and 30° respectively.
Let AD = x m and CD = h m
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 21.1
In right ∆ ACD,
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 21.2
Height of the tree = 17.32 m

Question 22:
Let AB be the building 7 meters high. AE ⊥ CD, where CD is the cable tower.
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 22.1
In ∆ AED,
∠EAD = 30° = Angle of depression
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 22.2
Height of the tower = CD = CE + ED = (21 + 7) m = 28 m

Question 23:
Let AB be the tower and let C and D be the two positions of the observer. Then, AC = 9 meters, and AD = 4 meters.
Let ∆ ACB = θ
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 23.1
Then, ∠ADB = (90° – θ)
Let AB = h meters
From right ∆ CAB, we have
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 23.2
From right ∆ DAB, we have
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 23.3
Hence, the height of tower is 6 meters.

Question 24:
Let P be the point of observation RQ is the building and BR is the flag staff of height h, ∠BPQ = 45°, ∠RPQ = 30°
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 24.1
From (1) and (2), we have
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 24.2
Hence distance of building is and length of the flags staff is 7.3 m

Question 25:
Let AB be the 10 m high building and let CD be the multi – storey building. Draw BE ⊥ CD
Then, ∠DBE = 30° and ∠DAC = 45°
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 25.1
Let ED = x meters
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 25.2
Height of the Multi – storey building = (10 + 13.66)m = 23.66 m
Distance between two building = (10 + 13.66) m = 23.66 m

Question 26:
Let A and B be two points on the bank on opposite sides of the river. Let P be a point on the bridge at a height of 2.5 m
Thus, DP = 2.5 m
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 26.1
Then, ∠BAP = 30°, ∠ABP = 45° and PD = 2.5m
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 26.2
Height of the river = AB
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 26.3

Question 27:
Let AB be the tower. Let C and D be the positions of the two men.
Then, ∠ACB = 30°, ∠ADB = 45° and AB = 50 m
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 27.1

Question 28:
Let AB and CD be the first and second towers respectively.
Then, CD = 90 m and AC = 60 m.
Let DE be the horizontal line through D.
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 28.1
Draw BF ⊥ CD,
Then, BF = AC = 60 m
∠FBD = ∠EDB = 30°
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance 14 28.2

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