Question

The angle of elevation of a tower at a point is 45°. After going 40 m towards the foot of the tower, the angle of elevation of the tower becomes 60°.Find the height of the tower.

Answer 1

LINE OF SIGHT: The line of sight is a line drawn from the eye of an observer to the point in the object viewed by the observer.

ANGLE OF ELEVATION: The angle of elevation of an object viewed is the angle formed by the line of sight with the horizontal , when it is above the horizontal level.

ANGLE OF DEPRESSION:The angle of depression of an object viewed is the angle formed by the line of sight with the horizontal , when it is below the horizontal level.

•Angle of elevation and depression are always acute angles.

•If the observer moves towards the perpendicular line(Tower/ building) then angle of elevation increases and if the observer move away from the perpendicular line(Tower/ building) angle of elevation decreases.

SOLUTION:

GIVEN:
∠ADB = 60° , ∠ACB = 45° , CD= 40 m, BD= x m  

Let AB = h m be the height of the tower.

In ∆ ABD
tan 60° = AB/BD = P /B
√3 = h /x
h = x √3
x = h/√3 ………………………... (1)

In ∆ ABC
tan 45° = AB/BC
tan 45° = AB/(BD+DC)
1 = AB/(x+40)
1= h /(x+40)
h = x + 40 ………………………. (2)

Put the value of x from eq 1 in eq 2.

h = x + 40
h = h/√3 +40
h - h/ √3  = 40
(√3h -h ) /√3 = 40
h (√3 -1) = 40√3
h = 40√3 /(√3 -1)
h = [40√3 /(√3 - 1)] x [(√3 +1)/(√3 +1)]

[By rationalising]
h = [120 + 40√3] / [(√3)² - 1²]
h = [120 + 40× 1.732] / 2
h  = (120 + 69.28)/2 = 189.28/2 = 94.64 m

Hence, the height of the tower is 94.64 m.

HOPE THIS WILL HELP YOU..

Answer 2

your answer is in the attachment