The 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 degrees find the height of the tower and the distance of the tower from the point A.

From ΔABC:

tan (θ) = opp/adj

tan (30) = BC/AB

BC = AB tan (30)—–(1)

From ΔBCD:

tan (θ) = opp/adj

tan (60) = BC/BD

BC = BD tan (60)——(2)

Equate (1) and (2)

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

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