Class 10th. 2018 1 37he angle of elevation of the top of a tower from

1.	Class 10th. 2018 1 37he angle of elevation of the top of a tower from a point on a ground level is0°. If on walking 20 metres towards the tower the angle of elevation of the topbecomes45, then the height of the towerOr
Answer» 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

Class 10th. 2018 1 37he angle of elevation of the top of a tower from a point on a ground level is0°. If on walking 20 metres towards the tower the angle of elevation of the topbecomes45, then the height of the towerOr

Answer»

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