Class 10 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 14 - Heights and Distances
Heights and Distances Exercise Ex. 14
Solution 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°
Let AB = h meters
From the right 
OAB, we have
Hence the height of the tower is 
Solution 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
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
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
In right 
ABS,
In right PAB,
Thus, height of the pedestal = 2m
Solution 9
Let AB be the unfinished tower and let AC be complete tower.
Let O be the point of observation. Then,
OA = 75 m
AOB = 30° and AOC = 60°
Let AB = h meters
And AC = H meters
Hence, the required height is 
Solution 10
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°
From right angled 
BOA
From right angled OAC
Thus 
Hence, height of the tower= 5.196 m and the height of the flagpole = 10.392 m
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
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.
Draw BF 
CD,
Then, BF = AC = 60 m
FBD = EDB = 30°
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Let AB be the height of the deck and let CD be the cliff..
Let the man be at B, then, AB= 16 m
Let BE 
CD and AE CD
Then, 
EBD = 60 and EBC = 30
CE = AB = 16m
Let CD = h meters
Then, ED = (h 16)m
From right 
BED, we have
From right CAB, we have
Hence the height of cliff is 64 m and the distance between the cliff and the ship = 
Solution 24
Solution 25
Solution 26
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
In 
CAB, we have
Hence the height of tower is 129.9 m
Solution 27
Let AB be the light house and let C and D be the positions of the ship.
Llet AD =x, CD = y
In 
BDA,
The distance travelled by the ship during the period of observation = 115.46 m
Solution 28
Solution 29
Solution 30
Solution 31
Solution 32
Solution 33
Solution 34
Let AB be the tower, and car moves from point D to C.
Now, we have
In ΔABC
Also, we have
In ΔABD
Now,
Solution 35
Let A be the aero plane,
BD be the river
AC is the perpendicular height = 300m
Now, we have
In ΔABC
Also, we have
In ΔADC
Solution 36
Let AD be the tower,
B be the point on ground,
CD will be the building.
Now, we have
In ΔBCD
Also, we have
In ΔABC
Solution 37
AB is the hill
C and D are the position of kilometer stones
Now, we have
In ΔABC
Also, we have
In ΔABD
Hence, height of hill is 1366m.
Solution 38
Let,
AC is the height of pole
BC is the shadow
In ΔABC
The angle of elevation of the sun is 60o.
Solution 39
AB is the tower
C and D are positions of boat
Now, we have
In ΔABC
Also, we have
In ΔABD
Hence the speed of the boat is 57.73 m/min.
Solution 40
AD be the multistoried building
BE be the small building
Now, we have
In ΔADE
Also, we have
In ΔABC
Hence the height of the multistoried building = 10.93+8 = 18.928 m
And the distance between both buildings is also 18.928 m
Heights and Distances Exercise MCQ
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
