RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance


Here you can get solutions of RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance. These Solutions are part of RS Aggarwal Solutions Class 10. we have given RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance download pdf.





RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance





Exercise 14





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-class-10-solutions-height-and-distance-14-q1-1
Let AB = h meters
From the right ∆OAB, we have
rs-aggarwal-class-10-solutions-height-and-distance-14-q1-2
Hence the height of the tower is [latex]20sqrt { 3 } m=34.64m    [/latex]





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-class-10-solutions-height-and-distance-14-q2





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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 ex 14
In ∆ACE,
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance ex 14
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 ex 14
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 Class 10 Solutions Chapter 14 Height and Distance ex 14
From right ∆ECD, we get
RS Aggarwal Class 10 Solutions Chapter 14 Height and Distance ex 14
From right ∆EAB, we get
RS Aggarwal Class 10 Solutions Chapter 14 Height and Distance ex 14
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
Solution of RS Aggarwal Class 10 Chapter 14 Height and Distance ex 14
From right ∆BCD, we have
Solution of RS Aggarwal Class 10 Chapter 14 Height and Distance ex 14
From (1) and (2), we get
Solution of RS Aggarwal Class 10 Chapter 14 Height and Distance ex 14
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 2017 Chapter 14 Height and Distance ex 14
From ∆BCD
RS Aggarwal Solutions Class 10 2017 Chapter 14 Height and Distance ex 14
From right ∆CAD, we have
RS Aggarwal Solutions Class 10 2017 Chapter 14 Height and Distance ex 14
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
Class 10 RS Aggarwal Solutions Chapter 14 Height and Distance ex 14
In CAE,
∠CAE = 30°and ∠AEC = 90°
Class 10 RS Aggarwal Solutions Chapter 14 Height and Distance ex 14
In CBD, ∠CBD = 60° and ∠CDB = 90°
Class 10 RS Aggarwal Solutions Chapter 14 Height and Distance ex 14
Eliminating x from (1) and (2), we get
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 14 Height and Distance ex 14
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 Class 10 Maths Book Solutions pdf Chapter 14 Height and Distance ex 14
AE is drawn perpendicular to CD
Therefore, AE = BD = 15 m
Let CE = h m
In ∆ ACE,
∠CAE = 30°, ∠CEA = 90°
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 14 Height and Distance ex 14
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 Class 10 Book pdf download Chapter 14 Height and Distance ex 14
In ∆ PBC,
∠CPB = 60° and ∠CBP = 90°
RS Aggarwal Class 10 Book pdf download Chapter 14 Height and Distance ex 14
Putting value of h in (1), we get
RS Aggarwal Class 10 Book pdf download Chapter 14 Height and Distance ex 14
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
Class 10 Maths RS Aggarwal Solutions Chapter 14 Height and Distance ex 14
In ∆ PBC,
∠CPB = 60°, ∠PBC = 90°
Class 10 Maths RS Aggarwal Solutions Chapter 14 Height and Distance ex 14
In ∆ APB,
∠APB = 30° and ∠ABP = 90°
Class 10 Maths RS Aggarwal Solutions Chapter 14 Height and Distance ex 14
Putting value of x in (1), we get
Maths RS Aggarwal Class 10 Solutions Chapter 14 Height and Distance ex 14
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
Maths RS Aggarwal Class 10 Solutions Chapter 14 Height and Distance ex 14
In right ∆ ABS,
Maths RS Aggarwal Class 10 Solutions Chapter 14 Height and Distance ex 14
In right ∆ PAB,
RS Aggarwal Solutions Class 10 2018 Chapter 14 Height and Distance ex 14
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 2018 Chapter 14 Height and Distance ex 14
In ∆ CAB, we have
RS Aggarwal Solutions Class 10 2018 Chapter 14 Height and Distance ex 14
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 ex 14
From right angled ∆ BOA
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance ex 14
From right angled ∆ OAC
RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance ex 14
Thus RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance ex 14
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 ex 14
Height of the pillar = CD = 133.33 m
Distance of the pillar from the hill = ED = [latex]frac { { 200 } }{ { sqrt { 3 }  } } times frac { { sqrt { 3 }  } }{ { sqrt { 3 }  } }     [/latex] = 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 Class 10 Solutions Chapter 14 Height and Distance ex 14
From right ∆ CAB, we have
RS Aggarwal Class 10 Solutions Chapter 14 Height and Distance ex 14
From right ∆ BED, we have
RS Aggarwal Class 10 Solutions Chapter 14 Height and Distance ex 14
Hence, the height of the opposite house is [latex]60(1+sqrt { 3 } )    [/latex]





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 Class 10 Solutions Chapter 14 Height and Distance ex 14
From right ∆ OCA, we have
RS Aggarwal Class 10 Solutions Chapter 14 Height and Distance ex 14
From right ∆ ADB, we have
rs-aggarwal-class-10-solutions-height-and-distance-14-q17-3
Thus, the aeroplane covers 3000 m in 15 seconds
Hence the speed of the aeroplane is rs-aggarwal-class-10-solutions-height-and-distance-14-q17-4





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
Solution of RS Aggarwal Class 10 Chapter 14 Height and Distance ex 14
In right ∆ ACE, let CE = h
Solution of RS Aggarwal Class 10 Chapter 14 Height and Distance ex 14
From (1) and (2),
[latex]20sqrt { 3 } =sqrt { 3 } h [/latex]
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
Solution of RS Aggarwal Class 10 Chapter 14 Height and Distance ex 14
In ∆ BDA,
RS Aggarwal Solutions Class 10 2017 Chapter 14 Height and Distance ex 14
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 2017 Chapter 14 Height and Distance ex 14
From right ∆ CAD, we have
RS Aggarwal Solutions Class 10 2017 Chapter 14 Height and Distance ex 14
From right ∆ BCD, we have
Class 10 RS Aggarwal Solutions Chapter 14 Height and Distance ex 14
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
Class 10 RS Aggarwal Solutions Chapter 14 Height and Distance ex 14
In right ∆ ACD,
Class 10 RS Aggarwal Solutions Chapter 14 Height and Distance ex 14
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.
Class 10 RS Aggarwal Solutions Chapter 14 Height and Distance ex 14
In ∆ AED,
∠EAD = 30° = Angle of depression
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 14 Height and Distance ex 14
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 Class 10 Maths Book Solutions pdf Chapter 14 Height and Distance ex 14
Then, ∠ADB = (90° – θ)
Let AB = h meters
From right ∆ CAB, we have
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 14 Height and Distance ex 14
From right ∆ DAB, we have
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 14 Height and Distance ex 14
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 Class 10 Maths Book Solutions pdf Chapter 14 Height and Distance ex 14
From (1) and (2), we have
RS Aggarwal Class 10 Book pdf download Chapter 14 Height and Distance ex 14
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 Class 10 Book pdf download Chapter 14 Height and Distance ex 14
Let ED = x meters
RS Aggarwal Class 10 Book pdf download Chapter 14 Height and Distance ex 14
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
Class 10 Maths RS Aggarwal Solutions Chapter 14 Height and Distance ex 14
Then, ∠BAP = 30°, ∠ABP = 45° and PD = 2.5m
Class 10 Maths RS Aggarwal Solutions Chapter 14 Height and Distance ex 14
Height of the river = AB
Class 10 Maths RS Aggarwal Solutions Chapter 14 Height and Distance ex 14





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
Class 10 Maths RS Aggarwal Solutions Chapter 14 Height and Distance ex 14





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.
Maths RS Aggarwal Class 10 Solutions Chapter 14 Height and Distance ex 14
Draw BF ⊥ CD,
Then, BF = AC = 60 m
∠FBD = ∠EDB = 30°
Maths RS Aggarwal Class 10 Solutions Chapter 14 Height and Distance ex 14





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