Question

From a point P on the ground the angle of elevation of the top of a tower is 30° and that of the top of a flag staff fixed on the top of the tower is 60°. If the length of the flag staff is 5 m , them find the height of the tower.​

Answer 1

✬ Height = 2.5 m ✬

Step-by-step explanation:

Given:

• Angle of elevation of top of tower is 30°.
• Angle of elevation of flag fixed on top of tower is 60°.
• Length of flag staff is 5 m.

To Find:

• What is the height of tower ?

Solution: Let AD be a flag fixed on the top of tower i.e DB. Let height of tower (DB) be x m.

Now in ∆DBC we have

• DB = Perpendicular = x m
• BC = Base
• ∠BCD = θ = 30°

Using tanθ in ∆DBC

• tanθ = Perpendicular/Base

➯ tan30 = DB/BC

➯ 1/√3 = x/BC

➯ BC = √3x.......(eqⁿ 1)

Now in ∆ABC we have

• AB = AD + DB = Perpendicular
• BC = Base
• ∠ACB = θ = 60°

Using tanθ in ∆ABC

$\implies{\rm }$ tanθ = p/b

$\implies{\rm }$ tan60° = AB/BC

$\implies{\rm }$ √3 = AD + DB/√3x

$\implies{\rm }$ √3 = 5 + x/√3x

$\implies{\rm }$ √3x × √3 = 5 + x

$\implies{\rm }$ 3x = 5 + x

$\implies{\rm }$ 3x – x = 5

$\implies{\rm }$ 2x = 5

$\implies{\rm }$ x = 5/2 = 2.5

Hence, height of the tower is 2.5 m.

Answer 2

Answer:

✡ Given ✡

• ✏ From a point P on the ground the angle of elevation of the top of a tower is 30°.
• ✏ The top of a flag staff fixed on the top of the tower is 60°.
• ✏ The length of flag staff is 5m.

✡ To Find ✡

✏ What is the height of the tower.

✡ Solution ✡

➡ Let AB denotes the height of the tower and BC denotes the height of the flag.

$\leadsto$ tan 30° = $\dfrac{AB}{AP}$

$\implies$ AP = √3AB ......(1)

$\leadsto$ tan 60° = $\dfrac{AC}{AP}$

$\implies$ AP = $\dfrac{1}{√3}$ (AB + 5) ....... (2)

▶ From (1) and (2) we get,

$\implies$ $\dfrac{1}{√3}$ (AB +5) = √3 AB

$\implies$ 3AB = AB + 5

$\implies$ 2AB = 5

$\implies$ AB = $\dfrac{5}{2}$

$\implies$ AB = 2.5 m

$\therefore$ Height of the tower is 2.5 m.

Step-by-step explanation:

HOPE IT HELP YOU