The shadow of a tower standing on a level plane is found to be 40 m longer when the sun 's altitude is `45^(@)` , than when it is `60^(@)` . The height of the tower is

Options

A. 644 m
B. 66.644 m
C. 54.644 m
D. 76.644 m

Answer

C

Explanation

Given, DC = 40 m , ` angle ACB = 30^(@) and angle ADB = 45^(@)`
In ` Delta ABD `
` tan 45^(@) = (AB)/( BD)`
`:. AB = BD " "` [`:. tan 45^(@) = 1`] . . . (i)
In ` Delta ABC `
`tan 30^(@) = (AB)/(BC) = (AB)/(BD + CD)`
`rArr (1)/( sqrt(3)) = (AB)/( BD + DC)`
`rArr sqrt(3) AB = BD + DC ` . . . (ii)

On putting the value of Eq. (i) in Eq. (ii) we get
` sqrt(3) AB = AB + 40`
`rArr sqrt(3) AB - AB = 40`
`rArr AB ( sqrt(3) - 1) = 40 " "` [`:. sqrt(3) = 1.732`]
`rArr AB = (40)/( sqrt(3) - 1) = (40)/(0.732)`
`= 54.644` m