Question

An aeroplane is flying parallel to the Earth’s surface at a speed of 175 m/sec and at a height of 600 m. The angle of elevation of the aeroplane from a point on the Earth’s surface is 37°. After what period of time does the angle of elevation increase to 53°? (tan 53° = 1.3270, tan 37° = 0.7536)

Answer 1

Let C is the initial and D is the final position of the aeroplane.

Let the time taken by the aeroplane be t

∴ CD = 175 t  ...(Distance = speed × time)

Let AB be x

∴ AE = x + 175 t

In the right ∆ABC

tan 53° = `"BC"/"AB"`

⇒ 1.3270 = `600/x`

x = `600/1.327`   ...(1)

In the right ΔAED, tan 37° = `"DE"/"AE"`

0.7536 = `600/(x + 175"t")`

x + 175 t = `600/(0.7536)`

x = `600/(0.7536) - 175"t"`  ...(2)

From (1) and (2) we get

`600/(1.327) = 600/(0.7536) - 175"t"`

175 t = `600/(0.7536) - 600/(1.327)`

175 t = 796.18 − 452.15 = 344.03

∴ t = `(344.03)/175` = 1.97 seconds

∴ Time taken is 1.97 seconds