Harsha is riding vertically in a hot air balloon, directly over a point A on the ground. Harsha spots a parked car on the ground at an angle of depression of 30°. The balloon rises 50 meters. Now the angle of depression to the car is 60 degrees. How far is the car from point A?
Explanation:
Given, BD = 50 m.
Let AC = x.
In ∆CAB, tan30° = 1/√3 = AB/AC
⇒ AB = x/√3 = x/√3 …(1)
In ∆CAD, tan60° = √3 = AD/AC
⇒ AD = √3 × x = √3 × x …(2)
We know, AD – AB = BD = 50
From (1) and (2)
⇒ √3x – x/√3 = 50
⇒ 2x/√3 = 50
⇒ x = 25√3.
Hence, option (b).
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