While landing karan airport, a pilot made an angle of depression of 20°. Average speed of the plane was 200 km/hr. The plane reached the ground after 54 seconds. Find the height at which the plane was when it started landing. (sin20° = 0.342)
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Answer:
Take angle of dispersion as given
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Step-by-step explanation:
let point 'A' be the position of plane
'AB' be the height of plane
angle of B = 90 degree
line AD is line of vision
angle DAC =angleACB. ( alternate angle)
angle ACB= 20 degree
In right angled triangle ABC
speed =
200km/hr = AC ÷ 54/3600. (1hr = 3600 second)
200×54/3600 = AC
AC = 3km
sin 20 = opposite/ hypotenuse
0.342 =AB/AC
0.342 = AB/3
0.342×3 = AB
AB = 1.026 km
AB = 1026 m
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