Question

The angle of elevation of an aeroplane from a point P on the ground is 60°. After 15 seconds of flight, the angle of elevation changes to 30 degrees . If the aeroplane is flying at a speed of 720 km/hour, find its height.


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Answer 1

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

Given that,

The angle of elevation of an aeroplane from a point P on the ground is 60°.

After 15 seconds of flight, the angle of elevation changes to 30 degrees .

The aeroplane is flying at a speed of 720 km/hour.

Let assume that the aeroplane is at the height of h m from the ground level.

So, QR = ST = h m

As it is given that,

$\rm \: Speed \: of \: aeroplane \: = \: 720 \: km \: per \: hour$

$\rm \: = \:720 \times \dfrac{5}{18} \: m \: per \: sec$

$\rm \: = \:200 \: m \: per \: sec$

So, Distance covered by plane during 15 seconds is

$\rm \: = \:15 \times 200 \:$

$\rm \: = \:3000 \: m$

So, RS = QT = 3000 m

Now, from the figure,

In triangle PQR

$\rm \: tan60 \degree \: = \: \dfrac{QR}{PQ}$

$\rm \: \sqrt{3} \: = \: \dfrac{h}{x}$

$\rm\implies \:h \: = \: \sqrt{3}x - - - (1)$

Now, In triangle PTS

$\rm \: tan30 \degree \: = \: \dfrac{TS}{PT}$

$\rm \: \dfrac{1}{ \sqrt{3} } \: = \: \dfrac{h}{PQ + QT}$

$\rm \: \dfrac{1}{ \sqrt{3} } \: = \: \dfrac{ \sqrt{3} x}{x + 3000}$

$\rm \: x + 3000 = 3x$

$\rm \: 2x = 3000$

$\rm\implies \:x \: = \: 1500 \: m$

On substituting the value of x, in equation (1), we get

$\rm\implies \:h \: = \: 1500 \sqrt{3} \: m$

So,

$\rm\implies \:Height \: of \: the \: aeroplane \: = \: 1500 \sqrt{3} \: m$

$\rule{190pt}{2pt}$

Additional Information :-

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