Question

An observer 1.4 m tall is 28.6 m away from a tower 30 m high. The angle of elevation of the top of the tower from his eye is

Answer 1

Answer:

45°

Step-by-step explanation:

From figure :

Let the observer be at AE = 1.4 m.

Let BD be the tower = 30 m.

∠BAC = ∅, BC = 30 - 1.4 = 28.6 m.

In ΔBAC,

tan∅ = BC/AC

⇒ tan∅ = (28.6)/(28.6)

⇒ tan∅ = 1

⇒ tan∅ = tan 45°

⇒ ∅ = 45°

Therefore,

angle of elevation = 45°

Hope it helps!

Answer 2

$\huge{\underline{\underline{\purple{\sf{Answer :}}}}}$

★ Given :

Height of tower (EC) = 30 m

Distance of observer from the tower (BC)= 28.6 m

Height of observer (AB) = 1.4 m

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★ To Find :

Angle of elevation of the top of the tower from the eye of the observer

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★ Solution :

A.T.Q

We have to find, Angle of elevation from the eye of the observer. So,

EC - CD(AB) = DE => 28.6 m, So that we can find the angle of elevation from the eye of observer.

_________________

Now

Let the angle be x.

We know that,

$\Large{\boxed{\boxed{\green{\sf{tan \: x = \frac{Perpendicular}{Base}}}}}}$

Where,

Base = AD = 28.6 m

Perpendicular = DE = 28.6 m

(Putting Values)

tan x = 28.6/28.6

tan x = 1

We know that,

$\large{\boxed{\blue{\sf{tan \: 45^{\circ} = 1}}}}$

So,

tan x = tan 45°

x = 45°

$\rule{200}{2}$

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