Question

the angle of elevation of an aeroplane from a point on the ground is 60 degree after 15 seconds later the elevation changes to 30 degree if the aeroplane is flying at a height of 1500√3 metre . find the speed of the plane


plz answer correctly.....
10 marks for each who answer
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Answer 1

Answer:

Step-by-step explanation:

 (Draw diagram based on gives conditions. mark E , D,C  points down and A,B points up.)

Tan 60 = AD/ED

= 1500/ED

ED = 1500

Tan 30 = BC/EC

1/ = 1500√3/EC

DE = EC - ED

4500-1500

= 3000

PLANE TRAVELS 3000 m  in 15 sec.

speed = Dstance/Time

= 3000/15

200 m/s.

Jet plane speed = 200m/s.

Answer 2

AnsweR :

Speed = 200m/s.

$\bf{\Large{\underline{\sf{Given\::}}}}$

The angle of elevation of an aeroplane from a point on the ground is 60° after 15 seconds later the elevation changes to 30°, if the aeroplane is flying at a height of 1500√3 m.

$\bf{\Large{\underline{\sf{To\:find\::}}}}$

The speed of the plane.

$\bf{\Large{\underline{\rm{\orange{Explanation\::}}}}}$

Let the speed of the Plane be MN.

We have,

Height of aeroplane = 1500√3 m.

$\leadsto\sf{PM=QN=h=1500\sqrt{3} m}\\\\\leadsto\sf{PQ=MN \:\:\:\:\:\:\:\:\:\bigg[Property\:of\:Parallelogram\:\bigg]}$

A/q

$\bigstar\bf{\large{\underline{\sf{In\:right\:angled\:\triangle\:APM\::}}}}$

$\longrightarrow\sf{tan60\degree\:=\:\dfrac{Perpendicular}{Base} }\\\\\\\\\longrightarrow\sf{tan60\degree\:=\:\dfrac{PM}{AM} }\\\\\\\\\longrightarrow\sf{\sqrt{3} \:=\:\dfrac{1500\sqrt{3} }{AM} }\\\\\\\\\longrightarrow\sf{\sqrt{3} AM\:=\:1500\sqrt{3} }\\\\\\\\\longrightarrow\sf{AM\:=\:\dfrac{1500\cancel{\sqrt{3}} }{\cancel{\sqrt{3} }} }\\\\\\\\\longrightarrow\sf{\pink{AM\:=\:1500\:m}}$

$\bigstar\bf{\large{\underline{\sf{In\:right\:angled\:\triangle\:AQN\::}}}}$

$\longrightarrow\sf{tan30\degree\:=\:\dfrac{QN}{AN} }\\\\\\\\\longrightarrow\sf{\dfrac{1}{3} \:=\:\dfrac{1500\sqrt{3} }{AN} }\\\\\\\\\longrightarrow\sf{AN\:=\:1500\sqrt{3} *\sqrt{3} }\\\\\\\\\longrightarrow\sf{AN\:=\:(1500*3)m}\\\\\\\\\longrightarrow\sf{\pink{AN\:=\:4500\:m}}$

Now,

$\mapsto\sf{AM+MN=AN}\\\\\\\mapsto\sf{1500m+MN=4500m}\\\\\\\mapsto\sf{MN\:=\:(4500-1500)\:m}\\\\\\\mapsto\sf{\pink{MN\:=\:3000\:m}}$

Hence,

Total Distance = 3000 m

Formula use :

$\bf{\large{\boxed{\rm{Speed=\frac{Distance}{Time} }}}}}$

$\longmapsto\sf{Speed\:=\:\cancel{\dfrac{3000}{15}} }\\\\\\\\\longmapsto\sf{\pink{Speed\:=\:200m/sec.}}$

Thus,

The speed of an aeroplane is 200m/sec.