Question

the angle of elevation of an airplane from a point of the ground us 60° after flying for 30sec the angle of elevation changes to 30° if the airplane is flying at a height of 4500m find the speed of the airplan
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Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that A be the point on the ground and Let plane at point B when angle of elevation of plane from A is 60°. Let further assume that plane at point C, after 30 seconds and angle of elevation of plane from point A is 30°.

Now, Further given that plane is flying at a height of 4500 m from the ground.

So, BD = CE = 4500 m

Time of flight during BC = 30 seconds.

Now, In triangle ABD

$\rm :\longmapsto\:tan60 \degree \: = \: \dfrac{BD}{AD}$

$\rm :\longmapsto\: \sqrt{3} \: = \: \dfrac{4500}{AD}$

$\rm :\longmapsto\: AD\: = \: \dfrac{4500}{ \sqrt{3} }$

$\rm :\longmapsto\: AD\: = \: \dfrac{4500}{ \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} }$

$\rm :\longmapsto\: AD\: = \: \dfrac{4500 \sqrt{3} }{3}$

$\bf\implies \:\boxed{\tt{ \: \: AD = 1500 \sqrt{3} \: m \: \: }} - - - - (1)$

In triangle ACE

$\rm :\longmapsto\:tan30 \degree \: = \: \dfrac{CE}{AE}$

$\rm :\longmapsto\: \dfrac{1}{ \sqrt{3} } \: = \: \dfrac{4500}{AE}$

$\rm\implies \:\boxed{\tt{ \: \: AE \: = \: 4500 \sqrt{3} \: m \: \: }} - - - - (2)$

Now,

$\rm :\longmapsto\:BC = DE = AE - AD = 4500 \sqrt{3} - 1500 \sqrt{3} = 3000 \sqrt{3} \: m$

So, Speed of the aeroplane during 30 seconds is

$\rm :\longmapsto\:Speed = \dfrac{3000 \sqrt{3} }{30}$

$\bf\implies \:Speed = 100 \sqrt{3} \: m \: per \: second$

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$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

Let assume that A be the point on the ground and Let plane at point B when angle of elevation of plane from A is 60°. Let further assume that plane at point C, after 30 seconds and angle of elevation of plane from point A is 30°.

Now, Further given that plane is flying at a height of 4500 m from the ground.

So, BD = CE = 4500 m

Time of flight during BC = 30 seconds.

Now, In triangle ABD

$\rm :\longmapsto\:tan60 \degree \: = \: \dfrac{BD}{AD}$

$\rm :\longmapsto\: \sqrt{3} \: = \: \dfrac{4500}{AD}$

$\rm :\longmapsto\: AD\: = \: \dfrac{4500}{ \sqrt{3} }$

$\rm :\longmapsto\: AD\: = \: \dfrac{4500}{ \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} }$

$\rm :\longmapsto\: AD\: = \: \dfrac{4500 \sqrt{3} }{3}$

$\bf\implies \:\boxed{\tt{ \: \: AD = 1500 \sqrt{3} \: m \: \: }} - - - - (1)$

In triangle ACE

$\rm :\longmapsto\:tan30 \degree \: = \: \dfrac{CE}{AE}$

$\rm :\longmapsto\: \dfrac{1}{ \sqrt{3} } \: = \: \dfrac{4500}{AE}$

$\rm\implies \:\boxed{\tt{ \: \: AE \: = \: 4500 \sqrt{3} \: m \: \: }} - - - - (2)$

Now,

$\rm :\longmapsto\:BC = DE = AE - AD = 4500 \sqrt{3} - 1500 \sqrt{3} = 3000 \sqrt{3} \: m$

So, Speed of the aeroplane during 30 seconds is

$\rm :\longmapsto\:Speed = \dfrac{3000 \sqrt{3} }{30}$

$\bf\implies \:Speed = 100 \sqrt{3} \: m \: per \: second$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$