Question

The shadow of a tower standing on a level plane is found to be a 50m longer when sons elevation is 30 degree then when it is 60 degree find the height of the tower.

Answer 1

Answer:

        " Height of Tower = 25 √3 m or 43.30 m "

Step-by-step explanation:

Kindly see the Attachment.

In Δ ACD

tan θ = P/B

tan 30° = h / (50 + x)

1/√3 = h/(50 + x)

50 + x = √3 h

x - √3 h = - 50 ..... (i)

In Δ BCD

tan θ = P/B

tan 60° = h / x

√3 = h / x

√3 x = h

h = √3 x  .... (ii)   , Putting this value in (i) , we get

x - √3 ( √3 x ) = - 50

x - 3 x = - 50

- 2 x = - 50

    x = 25 m

Putting this value in (ii), we get

h = √3 ( 25 )

h = 25 √3 m  OR h = 43.30 m

       " Height of Tower = 25 √3 m or 43.30 m "

I hope it will help you.