Question

The angle of elevation of the top of a tower from a point on the ground ,which is 30m away frrom the foit of the tower is 30 degree is 30 degree.the height of tower is​

Answer 1

Answer:

10√3  m

Step-by-step explanation:

Let AB be the tower of height h meters and C be a point on the ground 30 meters away from foot of tower B, such that ∠ACB = 30°

Now,

∵ The tower is vertically h meters tall.

∴ ∠ ABC = 90°

So, Δ ABC is a right triangle.

In right triangle ABC,

tan 30° = AB / BC

⇒ 1/√3 = h / 30

⇒ h = 30 / √3

⇒ h = 10 * √3 * √3 / √3

⇒ h= 10√3 m.

So, height of the tower is 10√3 m.

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Answer 2

Step-by-step explanation:

In ∆ ABC,

tan ∅ = Perpendicular/Height

tan 30° = AB/BC

1/√3 = AB/30

30/√3 = AB

AB = 30/√3

Now, Multiplying numerator and denominator by 3 we get:

AB = 30/√3 × √3/√3

AB = 30√3/3

AB = 10√3

Therefore, the height of the tower is 10√3.