Question

The angle of elevation of an aeroplane from a point A on the ground is 60°. After a flight to 30 seconds, the angle of elevation changes to 30°. If the plane is flying at a constant height of
3600√3 meters, find the speed of the aeroplane.

Answer 1

$\boxed{ \boxed{ \overline {\underline{ \bf \red{SOLUTION ☻ }}}}}$

$\rm \: Let \: OX \: be \: the \: horizontal \: ground, \: A \: and \: B \: be \: the \: two \: positions$

$\rm \: of \: the \: plane \: and \: O \: be \: the \: point \: of \: observation.$

$\rm \: Here, \: AC = BD = 3600 \sqrt{3 \: } m$

$\angle \rm AOC = 60 \degree \: and \: \angle BOD = 30 \degree$

$\rm \: In \: right \: angled \: \triangle OCA,$

$\rm \: cot \: 60 \degree = \frac{OC}{AC}$

$\longrightarrow \: \rm \frac{1}{ \sqrt{3} } = \frac{OC}{3600 \sqrt{3} }$

$\longrightarrow \: \rm \: OC = 3600 \: m$

$\rm \: In \: right \: angled \: \triangle ODB, \:$

$\rm \: cot \: 30 \degree \: = \frac{OD}{BD}$

$\longrightarrow \: \rm \: \sqrt{3} = \frac{OD}{3600 \sqrt{3} }$

$\longrightarrow \: \rm \: OD = 3600 \times 3 = 10800 \: m$

$\rm \: now \: CD = OD - OC = 10800 - 3600$

$\rm = 7200 \: m.$

$\rm \: Thus, \: distance \: covered \: by \: plane \: is \: 30 \: s \: is \: 7200 \: m.$

$\therefore \: \rm \: Speed \: of \: aeroplane = \frac{7200}{30} \times \frac{60 \times 60}{1000} = \red{864 \: km/h}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg( \because \rm \: Speed = \frac{distance}{time} \bigg)$

$\rm \: Hence, \: speed \: of \: aeroplane \: is \: \underline{ \underline{ \red{864 \: km/h.}}}$

Answer 2

Answer:

${ \purple{ \tan 60° = \frac{3000 \sqrt{3} }{x} }}$

${ \purple{x= \frac{3000 \sqrt{3} }{ \sqrt{3} } = 3000m}}$

${ \purple{ \tan 30° = \frac{3000 \sqrt{3} }{ \frac{1}{ \sqrt{3} } } = 9000m}}$

Distance covered :

${ \purple{ 9000 - 3000 = 6000m}}$

${ \purple{speed = \frac{6000}{30} }}$

${ \huge{ \boxed{ \purple{speed = 200 m/s}}}}$

Step-by-step explanation:

Hope this helps you ✌️