CBSE Class 10CBSE
Account
It's free!

User



Login
Register


      Forgot password?
Share
Notifications

View all notifications
Books Shortlist
Your shortlist is empty

Solution - The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the building - CBSE Class 10 - Mathematics

ConceptHeights and Distances

Question

The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the building (use `sqrt3`=1.73)

Solution

Let the height of the tower AB be h m and the horizontal distance between the tower and the building BC be x m.

So,

AE=(h50) m

In AED,

`tan45^@=(AE)/(ED)`

 `=>1=(h-50)/x`

 h50   .....(1)

 In ABC,

`tan^@=(AB)/(BC)`

`=>sqrt3=h/x`

`=>x=sqrt3=h `

Using (1) and (2), we get

`x=sqrt3x-50`

`=>x=(sqrt3-1)=50`

`=>x=(50(sqrt3+1))/2=25xx2.73=68.25m`

Substituting the value of x in (1), we get

68.25 50

⇒ 68.25 50

h=118.25 m

Hence, the height of tower is 118.25 m and the horizontal distance between the tower and the building is 68.25 m.

Is there an error in this question or solution?

APPEARS IN

2015-2016 (March) Delhi Set 1
Question 19 | 3 marks
2015-2016 (March) Delhi Set 2
Question 17 | 3 marks
2015-2016 (March) Delhi Set 3
Question 13 | 3 marks

Reference Material

Solution for question: The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the building concept: Heights and Distances. For the course CBSE
S