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The angle of elevation of an aeroplane from a point on the ground is 45°. After 15 seconds of flight, the elevation changes to 30°. If the aeroplane is flying at a height of 3000 m, the speed of the plane in km per hour is:

  1. 152.16

  2. 263.5

  3. 304.32

  4. 527

Heights & Distances

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Answer

Let A be the initial position of aeroplane and height AB = 3000 m, C be the final position and height CD = 3000 m.

O be the point of observation,

The angle of elevation of an aeroplane from a point on the ground is 45°. After 15 seconds of flight, the elevation changes to 30°. If the aeroplane is flying at a height of 3000 m, the speed of the plane in km per hour is: Volume And Surface Area of solid RSA Mathematics Solutions ICSE Class 10.

In triangle OBA,

⇒tan⁡45∘=ABOB⇒1=3000OB⇒OB=3000 m.\Rightarrow \tan 45^{\circ} = \dfrac{AB}{OB} \\[1em] \Rightarrow 1 = \dfrac{3000}{OB} \\[1em] \Rightarrow OB = 3000 \text{ m.}

In triangle COD,

⇒tan⁡(30∘)=CDOD⇒13=3000OD⇒OD=30003 m.\Rightarrow \tan (30^{\circ}) = \dfrac{CD}{OD} \\[1em] \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{3000}{OD} \\[1em] \Rightarrow OD = 3000 \sqrt3 \text{ m.}

The distance the plane flew is,

BD = OD - OB

BD = 3000 3\sqrt{3} - 3000

BD = 3000(3\sqrt{3} - 1)

BD = 3000(1.732 - 1) = 3000(0.732) = 2196 m.

⇒Speed=DistanceTime⇒Speed=219615⇒Speed=146.4 m/s⇒Speed=146.4×3.6⇒Speed=527.04≈527 km/h.\Rightarrow \text{Speed} = \dfrac{\text{Distance}}{\text{Time}} \\[1em] \Rightarrow \text{Speed} = \dfrac{2196}{15} \\[1em] \Rightarrow \text{Speed} = 146.4 \text{ m/s} \\[1em] \Rightarrow \text{Speed} = 146.4 \times 3.6 \\[1em] \Rightarrow \text{Speed} = 527.04 \approx 527 \text{ km/h}.

Hence, option 4 is the correct option.

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