Question

An observer 1.6 m tall is 203 away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The height of the tower is:

A. 21.6 m

B. 23.2 m

C. 24.72 m

D. None of these​

Answer 1

Step-by-step explanation:

Correct option is

A

21.6m

Let AB be the observer and CD tower

Draw BE perpendicular to CD

Then CE=AB=1.6 m

And BE=AC= 20

3

m

Then right angle triangle DEB

∴tan30

o

=

BE

DE

⇒

3

1

=

20

3

DE

⇒DE=20

3

m

Then CD=CE+DE=1.6+20=21.6 m

solution

Answer 2

Solution

Given :-

• An observer 1.6 m tall is 20√3 away from a tower.
• The angle of elevation from his eye ti the top of the tower is 30° .

Find :-

• Height of tower

Explantion

Let,

AB be observer and CD be the tower.

In graph .

• AB = CE = 1.6 m
• AC = EB = 20√3 m

And,

• <DBE = 30°

So, Now take right angle ∆DBE,

★ tan 30° = Perpendicular/Base

Where,

• Perpendicular = DE = h
• Base = BE = 20√3 m

Keep values,

==> tan 30° = h/20√3

We Know,

★ $\boxed{\tt{\red{\:\tan 30^{°}\:=\:\dfrac{1}{\sqrt{3}}}}}$

==> 1/√3 = h/20√3

==> h = 20√3 /√3

==> h = 20

==> h = 20 m

Now, Calculate CD

==> CD = CE + DE

==> CD = 1.6 + 20

==> CD = 21.6 m

Hence

• Height of tower will be( h) = 21.6 m

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