Question

the angle of elevation of the top of the tower from point A on the ground is 30 .on moving a distance of 20 m towards the foot of the tower to a point B .the angle of elevation increase to 60 find the hight of the tower and the distance of the tower from the point A (take root 3 = 1.732).

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

Hope it helps you dude......