Question

At point on ground, the angle of elevation of tower is 30°. On walking 30m towards tower, the angle of elevation becomes 60°. Find height of tower

Answer 1

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Let the tower as AB , the. point from where angle of elevation is 30° is C after walking 30m we reach at a point D here angle of elevation is 60° .

Let,

CB = x m

DB = x - 30 m

In ∆ ABD

tan 60° = AB / DB

√3. = h / x - 30

h. = √3x -30√3. (i)

In ∆ ABC

tan 30° = AB/ CB

1/√3. = h/ x

h. = x/√3. ( ii )

Putting eq 1 and 2 equivalent.

x/√3. = √3x - 30√3

x. = 3x - 90

2x. = 90

x. = 90/2

x. = 45 m .

putting value of X in eq 1

h = √3(45) - 30√3

h = √3 (45 - 30)

h = √3(15)

h = 15√3 [ √3 = 1.73 ]

h = 25.95 m

So, height of tower is 25.95m