ROUTERA


Chapter 14 Heights and Distances

Class 10th R. S. Aggarwal Maths Solution
CBSE Class 10 Maths
R. S. Aggarwal Solution


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

 

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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

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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°

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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 =

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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

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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

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