Question

angle of elevation of the top of a tower from a point a on the ground is 30 degree on moving a distance of 20 m towards the foot of the tower to a point B the angle of elevation increases to 60 degree find the height of the tower and the distance of the tower from the point A

Answer 1

From ΔABC:

tan (θ) = opp/adj

tan (30) = BC/AB

BC = AB tan (30)

From ΔBCD:

tan (θ) = opp/adj

tan (60) = BC/BD

BC = BD tan (60)

Equate the 2 equations:

AB tan (60) = BD tan (30)

Define x:

Let BD = x

AB = x + 20

Solve x:

AB tan (30) = BD tan (60)

(x + 20) tan (30) = x tan (60)

x tan (30) + 20 tan (30) = x tan (60)

x tan (60) - x tan (30) = 20 tan (30)

x ( tan (60) - tan (30) ) = 20 tan (30)

x = 20 tan (30) ÷ ( tan (60) - tan (60) )

x = 10 m

Find the distance:

Distance = 10 + 20 = 30 m

Find the height:

tan (θ) = opp/adj

tan (60) = BC/10

BC = 10 tan (60)  = 10√3 m

Answer:  Distance = 30 m and height = 10√3 m

Answer 2

Answer:

Height of the tower is 10√3

Step-by-step explanation:

Let BC Represent The Height of Tower

BC = h

Angle of elevation in Triangle ABC

= angle CAB = 30°

And after 20 m. towards angle of elevation is Angle CDB = 60°

In Triangle ABC

it is a right angle triangle

tan ∅ = opp/adj

tan ∅ = CB/ BA

1/√3 = h/x+20

X + 20 = h√3

X = h√3 -20

Now in Triangle BCD

it is also right angle triangle

tan ∅ = CB/BD

√3 = h/X

√3. = h/h√3 -20

√3 (h√3-20)= h

3h - 20√3 = h

-20h = -2h

h = 10√3

Height of the tower is 10√3.