Question

From a balloon vertically above a straight road, the angle of depression of two cars at an instant are found to be 30° and 45°. If the cars are 80 m apart, find the height of the balloon.

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Answer 1

Let the two cars be at points C and D. That is distance between the cars, CD = 100 m Let BD = d Hence BC = (100 – x) m Let the height of the balloon be AB = ‘h’ m In right ΔABD, θ = 45° In right ΔABC, θ = 60° ⇒ 100√3 – √3h = h ⇒ 100√3 = √3h + h = h(√3 + 1) ∴ h = 50√3(√3 + 1) m

Answer 2

Answer:

height = 109.28 m

Step-by-step explanation:

In ΔABC, ∠ACB = 30°

∴ tan 30° = AB/BC

=> 1/√3 = h/BC

=> BC = h√3    ----- (1)

In ΔABD, ∠ADB = 45°

∴ tan 45° = AB/BD

=> 1 = h/BD

=> BD = h m.

Points B,D and C are collinear.

BC = BD + DC

=> h√3 = h + 80

=> (h√3) - h = 80

=> h(√3 - 1) - h = 80

=> h = 40(1 + √3)

=> h = 40(1 + 1.73)

=> h = 109.28

Hence, height of the balloon = 109.28 m

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