Question

the angle of elevation of top of a tower from a point on the ground which is 30 metre away from the foot of the tower is 30 degree find the height of Tower

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:

Step-by-step explanatiaon:

Let the height of the tower be 'h'

Tan30° = 1/√3 = h/30

By cross multiplying,

h√3 = 30

h = 30/√3

By rationalising,

h =30√3/3

h = 10√3 m

Therefore the height of the tower is 10√3m