Question

a plane is observed to be approaching the airport it is the distance of 12 km from the point of the observation and makes an angle of elevation 60 the height above the ground ao the plane is

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Answer 1

Given :

• Angle of elevation = 60°

• Distance from the point of observation = 12 km

To find :

The height of the plane above the ground.

Solution :

Let the height of the plane be h km.

According to the diagram , we have to find the height (AB) and Base (BC) , so will use tan θ as :

$\bf{tan\:\theta = \dfrac{P}{B}}$

Where :-

• P = Height
• B = Base

Now by substituting the given values in tan θ, we get :

$:\implies \bf{tan\:\theta = \dfrac{h}{B}}$

$:\implies \bf{tan\:60^{\circ} = \dfrac{h}{12}}$

$:\implies \bf{\sqrt{3} = \dfrac{h}{12}}:\:\:\:\:[\because \bf{tan\:60^{\circ} = \sqrt{3}}]$

By Cross-multiplication , we get :-

$:\implies \bf{12\sqrt{3} = h}$

$:\implies \bf{12\sqrt{3} = h}$

$\underline{\therefore \bf{Height\:(h) = 12\sqrt{3}\:km}}$

Hence, the height of the plane above the ground is 12√3 km.

Answer 2

$\sf{ \pink{\Large{Given:}}}$

◎ Distance from the point of observation = 12 km

◎ An angle of elevation = 60°

$\sf{ \blue{\Large{To \: Find:}}}$

◍ The height above the ground of the plane.

$\sf{ \pink{\Large{Solution:}}}$

In ∆ ABC

∠ABC = 90°

In this question we have to find height above the ground, So

Let the height be h km.

Now, we know the distance from the point of observation i.e., Base is 12 km. For finding height h km we will tan θ.

We know,

$\sf\tan A = \dfrac{P}{B}$

Here,

P = Perpendicular

B = Base

Now we know it makes an angel of elevation of 60°. Therefore, by putting the value of 60° in tan A, we have:

$\sf \tan 60 \degree = \dfrac{BC}{AB}$

$\sf\tan 60 \degree = \dfrac{h}{12}$

Value of tan 60° = √3

$\sf\sqrt{3} = \dfrac{h}{12}$

On further solving:

$\sf{h = 12 \sqrt{ 3} }$

The height above the ground of the plane = $\green{ \underline{ \boxed{ \sf{ 12 \sqrt{3 \:} \: km}}}}$

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