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While landing at an airport, a pilot mad...

While landing at an airport, a pilot made an angle of derpession of `20^(@)`. Average speed of the plane was `200km//h`. The plane reached the ground after 54 seconds. Find the height at which the plane was when it started landing.
`(sin20^(@)=0.342)`

Text Solution

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Let A represent the position of the plane and point C represent the position where plane lands.
`AD` represents the horizontal.
`/_DAC` is the angle of depression
`/_DAC=20^(@)`
`/_ACB` and `/_DAC` are alternate angles.
`:./_ACB=20^(@)`
Average speed of the plane `=200km//h`
Time take by the plane from point A to `C=54` seconds
`=(54 "hours")/3600` [ `:'` 1 hour `=` 3600 seconds]
Distance `AC=` Speed `xx` Time
`:.AC=200xx54/3600`
`:.AC=54/18`
`:.AC=3km`
In right angled `DeltaABC`,
`sin 20=(AB)/(AC)` .........(By definition)
`:.0.342=(AB)/3 .......... ( :' sin 20^(2)=0.342)`
`:.AB=0.342xx3`
`:.AB=1.026km` or `1026m`
Plane was at a height of `1026 m` when it started landing.
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