Question

20. An aeroplane is flying horizontally at an
altitude of 1500 m above the ground. When
it is observed from a point on the ground, the
angle of elevation changes from 60° to 30° in
12 seconds. Find the speed of the plane in
km/h.
​

Answer 1

Answer:

250 m/sec

Step-by-step explanation:

speed =250m/sec

Answer 2

Given,

An airplane is at 1500m in height.

The first angle of elevation = 60°

The second angle of elevation = 30°

To Find,

The speed an airplane took to travel.

Solution,

Given that,

From the ground level, an airplane is at a height of 1500m or 15km.

The elevation angle,

∡PAB = 60°

∡QAC= 30°

From the figure, it's clear that PB and QC are the same.

So, PB= QC.

ABP is a right-angled triangle,

Therefore,

$\frac{BP}{AB}$ = tan60

we know the value of tan60 is $\sqrt{3}$

BP is 1.5 (1500m -> 1.5km)

$\frac{1.5}{AB}$ = $\sqrt{3}$

AB = $\frac{1.5}{\sqrt{3} }$

AB = 05 ×$\sqrt{3}$

Then in ΔACQ

tan30° = $\frac{QC}{AC}$

$\frac{1}{\sqrt{3} }$ = $\frac{1.5}{AC}$

Ac = 1.5×$\sqrt{3}$km.

Let's find AC - AB

1.5$\sqrt{3}$ - .5$\sqrt{3}$

$\sqrt{3}$(1.5 - .5)

$\sqrt{3}$km.

Travelling speed  = 12s.

Therefore,

speed = $\frac{\sqrt{3} }{t}$

time = $\frac{12}{3600}$

Therefore,

$\frac{\sqrt{3} }{\frac{12}{3600} }$ = 52km/h.

Hence, 520km/h is the speed airplane takes to travel.