Question

An aeroplane flying horizontally at a height of 2500m above the ground is observed at an elevation of 60 degree. If after 15 secs, the angle of elevation is observed to be 30 degree, find the speed of the aeroplane im km/hr. (Please give the appropriate figure too).

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$Speed \: of \: plane = 400\sqrt{3} \:km/hr$

Step-by-step explanation:

Let P and Q be the two positions of the plane and let A be the point of observation.

Let ABC be the horizontal line through A .

It is given that angles of elevation of the plane in two positions P and Q from a point A are 60° and 30° respectively.

$Then, \:\angle PAB =\angle 60\degree ,\\\angle QAB = 30\degree.\\and \: PB=2500\:m$

$In \triangle ABP, \:we \: have\\tan 60\degree = \frac{BP}{AB}\\\implies \sqrt{3}=\frac{2500}{AB}\\\implies AB = \frac{2500}{\sqrt{3}}$

$In \: \triangle ACQ ,\: we \: have \\tan\: 30\degree = \frac{CQ}{AC}\\\implies \frac{1}{\sqrt{3}}=\frac{2500}{AC}\\\implies AC=2500\sqrt{3} \:m$

$Therefore, \\PQ=BC=AC-AB\\=2500\sqrt{3}-\frac{2500}{\sqrt{3}}\\=\frac{2500\times 3 - 2500}{\sqrt{3}}\\=\frac{7500-2500}{\sqrt{3}}\\=\frac{5000}{\sqrt{3}}\: m$

$The \: plane \: travels \:\\ \frac{5000}{\sqrt{3}} \: m \: in \: 15 \: seconds .$

$Speed \: of \: plane \\=\frac{5000}{\sqrt{3}} \times \frac{1}{15}$

$=\frac{1000}{3\sqrt{3}}\: m/s$

$=\frac{1000}{3\sqrt{3}}\times \frac{3600}{1000}\: km/hr$

$=\frac{1200}{\sqrt{3}}\\=\frac{1200\sqrt{3}}{\sqrt{3}\times \sqrt{3}}\\=400\sqrt{3}\: km/hr$

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