Question

The angle of elevation of the top of a vertical tower from a point on the ground is 60 degree. From another point 10m vertically above the first, its angle of elevation is 45 degree. Find the height of the tower.

Answer 1

let height of tower = x + 10m
let dist. between base of tower and obs. pt. = y
at first angle it makes with ground = 60 degree
tan(60) = sqrt{3} = (x + 10)/y   . . . . . . (eq 1)
next angle = 45 degree
pt is 10 m above the ground 
tan(45)= 1 = x/y    
:. x=y  . . . . . . . . . (eq. 2)
substituting value of y in eq. 1 
x + 10 = 1.732 x
:. x = 13.66 m 

Answer 2

Answer:

23.66m

Step - by - Step  explanation:

Also according to the question the diagram can be drawn as in the pic.

So let height of the tower be EC

Also let the angle EAD be 45° and the angle EBC be 60°

Let ED be x and in the rectangle ABCD we know that AB = CD = 10

and BC = AD

Since we don't need hypotenuse value let's apply tan or cot.

( for easy simplification I am applying tan )

So, in triangle AED ,,,

Tan 45 ° = ED/AD = ED/BC = x / BC

1 = x/BC

x = BC ----------------------------(1)

In triangle EBC ,,,

Tan 60 ° = EC/BC

√3 = x + 10 / BC

BC(√3)  = x + 10

x√3 - x = 10

x ( √3 - 1) = 10

x = 10 / ( √3 - 1 )--------------------------(2)

According to the pic we can say that height of the tower is EC = ED + DC

                                          =  x + 10

                                          = [10/(√3-1)] +

                                                             10

                                         = (10+10√3-

                                               10)/(√3-1)

                                         = 10√3/(√3-1)

If we substitute √3 = 1.732

We get height if the tower as 23.66 m

Thanks

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