Question

An aeroplane is flying at a Constant height of 1200 root 3 meters The angle of elevation of that aeroplane from a point on the ground is 60⁰. After a flight 25 seconds , the angle of elevation changes to 30 ⁰ Find the speed of aeroplane

Answer 1

Answer:

In triangle ACE,

CE = 10800 m

AC = BD =

In triangle BED,

CD + DE = CE

CD + 3600 = 10800

CD = 10800 - 3600 = 7200 m

Distance travelled = 7200 m

Time taken = 30 seconds

In 1 second = 240 m

In 3600 seconds (1 hour) = 240 × 3600 = 864000 m = 864 km

In hour = 864 km

Speed = 864 km/h

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Answer 2

Answer:

An aeroplane is flying at a Constant height of 1200$\sqrt{3}$ meters The angle of elevation of that aeroplane from a point on the ground is 60⁰. After a flight 25 seconds , the angle of elevation changes to 30 ⁰ Find the speed of aeroplane?

Answer:  96 m/s

Step-by-step explanation:

Suppose speed of aeroplane is v m/s.

Given that constant height of aeroplane = BE = CE = 1200$\sqrt{3}$ meters

The angle of elevation of that aeroplane from a point on the ground is 60⁰.

∠EAB = 60°

After flight 25 seconds,

distance travelled by the aeroplane = 25v m

and angle of elevation = 30°

∠DAC = 30°

In ΔDAC

cot30° = AC / DC = AC / 1200$\sqrt{3}$

⇒ AC = 1200$\sqrt{3}$ Cot30°

Now, in Δ EAB

Cot60° = AB / EB = AB / 1200$\sqrt{3}$

⇒ AB = 1200$\sqrt{3}$ Cot60°

from figure,

BC = AC - AB

⇒25v = 1200$\sqrt{3}$ Cot30° - 1200$\sqrt{3}$ Cot60°

         = 1200$\sqrt{3}$ ( Cot30° - Cot60° )

         = 1200$\sqrt{3}$ ( $\sqrt{3}$ - 1 / $\sqrt{3}$ )

         = 1200$\sqrt{3}$ x 2 / $\sqrt{3}$

         = 2400

⇒ v   = 2400 / 25 = 96 m/s

Therefore the speed of aeroplane = 96 m/s