a diagram show two different point P and Q of a helicopter flying horizontally at the uniform speed of 72 km per hour it is found that it take the helicopter 10 minutes to fly from P to Q the angle of elevation of P and Q from the point A on the ground are 45 degree and 60 degree respectively find the distance PQ and the distance active and the height of the helicopter above the ground
Answers
Distance between P & Q = 12 km , height of the helicopter above the ground 18 + 6√3 km = 28.39 km
Step-by-step explanation:
Helicopter Speed = 72 km /hr
Distance covered in 10 mins = 72 * 10/60
= 12 km
Distance between P & Q = 12 km
Let Say Height of Helicopter above ground = H km
Tan 45 = Height of Helicopter above ground/AP
=> 1 = H/AP
=> AP = H
Tan 60 = Height of Helicopter above ground/AQ
=> √3 = H/AQ
=> AQ = H/√3
PQ = AP - AQ = 12 km
=> H - H/√3 = 12
=> H(√3 - 1) = 12√3
=> H = 12√3/(√3 - 1)
=> H = 12√3(√3 + 1)/(3 - 1)
=> H = 6(3 + √3)
=> H = 18 + 6√3 km
height of the helicopter above the ground 18 + 6√3 km = 28.39 km
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