Question

10. An aeroplane, when flying at a height of 4000 m from the ground passes
vertically above another aeroplane at an instant when the angles of elevation of
the two planes from then same point on the ground are 60° and 45° respectively.
Find the vertical distance between the aeroplane at that instant. [Take root3=1.73]​

Answer 1

Given :

• Height of the Aeroplane $\bf{(A_{1})}$ = 4000 m

• Angle of elevation $\bf{(v_{1})}$ = 60°

• Angle of elevation $\bf{(v_{2})}$ = 45°

To find :

The vertical distance between the two aeroplane at the instant.

Solution :

To find the vertical distance of the between the two instant first we need to find the height of the other aeroplane $\bf{(A_{2})}$.

According to the information , the Difference of the height of the two aeroplanes will give us the vertical distance between them.

But to find the distance of the second aeroplane , first we need to find the distance at the ground .

To find the distance of the shadow :-

According to the diagram , the Height (AB) is given and we have to find the Base (CB) , so we will use tan θ.

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

Now using the formula and substituting the given values in it , we get :

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

$:\implies \bf{tan\:60^{\circ} = \dfrac{4000}{B}}$

$:\implies \bf{\sqrt{3} = \dfrac{4000}{B}}\:\:\:\:[\because \bf{tan\:60^{\circ} = \sqrt{3}}]$

By cross-multiplication , we get :

$:\implies \bf{4000 = \sqrt{3}B}$

$:\implies \bf{\dfrac{4000}{\sqrt{3}} = B}$

$:\implies \bf{\dfrac{4000}{1.73} = B}\:\:\:\:[\because \bf{\sqrt{3} = 1.73}]$

$:\implies \bf{2312.1 = B}$

$\underline{\therefore \bf{Base\:(B) = 2312.1\:m}}$

Hence, the distance at the ground is 2312.1 .

To Find the height of the second Aeroplane :

According to the diagram , the Base (CB) is given and we have to find the Base (DB) , so we will use tan θ.

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

Now using the formula and substituting the given values in it , we get :

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

$:\implies \bf{tan\:45^{\circ} = \dfrac{P}{2312.1}}$

$:\implies \bf{1 = \dfrac{P}{2312.1}}\:\:\:\:[\because \bf{tan45^{\circ} = 1}]$

$:\implies \bf{1 = \dfrac{P}{2312.1}}$

By cross-multiplication , we get :

$:\implies \bf{1 \times 2312.1 = P}$

$:\implies \bf{2312.1 = P}$

$\underline{\therefore \bf{Height\:(P) = 2312.1\:m}}$

Hence, the height of the second aeroplane is 2312.1 m.

To find the Vertical distance between the two Aeroplanes :-

Here , let the height of AD be d.

So according to the given information , we get :

$\boxed{:\implies \bf{Height_{A_{1}} - Height_{A_{2}} = H_{d}}}$

$:\implies \bf{4000 - 2312.1 = H_{d}}$

$:\implies \bf{1687.9 = H_{d}}$

$\underline{\therefore \bf{Height\:(d) = 1687.9\:m}}$

Hence, the distance between two Aeroplanes at that instant is 1687.9 m.