Question

Q1.A man finds the angle of elevation of a tower to be 60°. When he walks away at a distance of 50m he finds the angle to be 30°, then the height of the tower is​?

Answer 1

Explanation:

✬ Height = 25√3 ✬

Step-by-step explanation:

Given:

Angle of elevation of tower is 60°.

After walking 50 m away from tower it changes to 30°.

To Find:

What is height of tower ?

Solution: Let the height of tower AB be h m and distance between the man initial position and foot of tower be x m. Therefore,

AB (perpendicular) = h

DB (base) = x

∠ABD = 90°

∠ADB (angle of elevation) = 60°

[ Now he walked 50 m away from point D. Let he is at now point C. ]

DC = 50 m

BC = DB + DC

∠ACB (angle of elevation) = 30°

In ∆ABD , using tanθ

★ tanθ = Perpendicular/Base ★

➯ tan60° = AB/DB

➯ √3 = h/x

➯ √3x = hㅤㅤㅤㅤㅤ(eqⁿ i)

Now in ∆ABC , again by tanθ

➯ tan30° = AB/BC

➯ 1/√3 = h/x + 50

➯ x + 50 = √3h

➯ x + 50 = √3 × √3x

➯ 50 = 3x – x

➯ 50/2 = x

➯ 25 m = x

Putting the value of x in eqⁿ (i)

$\implies{\rm }⟹ √3x = h$

$\implies{\rm }⟹ √3 × 25 = h$

$\implies{\rm }⟹ 25√3 = h$

Hence, the height of tower is 25√3 m.

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