Question

An aeroplane is flying horizontally along a straight line at a height of 3000 m from the ground at a speed of 160 m/s. Find the time it would take for the angle of elevation of the plane as seen from a particular point on the ground to change from 60⁰ to 45⁰. Give your answer correct to the nearest second.​

Answer 1

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Let height of first aeroplane is BD which is flying at 3000 m height and height of second aeroplane BC = hm . Let angles of elevation from point A are 60° and 45° respectively.

So, angle BAD=60° and angle CAB=45° and Let

AB = x m

From right angled ∆ABD,

tan 60° = (BD)/(AB)

√3=3000/x

From right angled ∆ABC

tan 45° = (BC)/(AB)

1 = h/x

1 = h/1000√3

h = 1000√3 = 1000 × 1.732

=1732m

Hence, height of second plane from first plane

CD=BD-BC=3000-1732=1268 m

Answer 2

Answer: 8 seconds

Step-by-step explanation:

In ▴ AEB,
Tan 60= EB/AB

Tan 60=3000/x

√3=3000/x {Tan 60= √3}

x=3000/√3 {multiplying numerator and denominator with √3}

Therefore, x=3000√3/3
x=1000√3

In ▴ ADC,

Tan 45=3000/AC

Tan 45=3000/AB+BC {let AB=x and BC be the distance travelled d= 160t}

Tan 45= 3000/x+160t

x+160t=3000 {Tan 45=1}
160t=3000-1000√3

160t=1000(3-√3)

160t=1000(3-1.732)

160t=1000×1.268

∴t=1268/160

t=7.925 seconds

but since we have to find time to the nearest second the answer will be

t=8 seconds

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