Question

: an aeroplane is moving one km high from west to easthorizontally from a point on the ground. The angle of elevation of the aeroplane is 60' and after 10 sec the angle of elevation of the plane is observed as 30'. Find the speed of the plane in km/hr

Answer 1

An aeroplane is moving one km high from west to east horizontally from a point on the ground. The angle of elevation of the aeroplane is 60° and after 10 sec the angle of elevation of the plane is observed as 30°. Find the speed of the plane in km/hr.

Solution:

Given:

$$A D = B C = 1 \mathrm {k m}, \angle A O D = 6 0 {}^ {\circ}, \angle B O D = 3 0 {}^ {\circ}$$

Let $x$ be the distance of the observer to the point on the ground at the beginning

$$x = O D$$

In $\Delta AOD\tan \angle AOD = \frac{AD}{OD}$ , so

$$O D = \frac {A D}{\tan \angle A O D} = \frac {1}{\tan 6 0 {}^ {\circ}} = 0. 5 8 \mathrm {k m}$$

Let $y$ be the distance of the observer to the point on the ground at the end

$$y = O C$$

In $\Delta BOC\tan \angle BOC = \frac{BC}{OC}$ , so

$$O C = \frac {B C}{\tan \angle B O C} = \frac {1}{\tan 3 0 {}^ {\circ}} = 1. 7 3 \mathrm {k m}$$

So the plane is flying for 10 seconds

$$y - x = 1. 7 3 - 0. 5 8 = 1. 1 5 \mathrm {k m}$$

The speed of the plane is

$$V = \frac {1 . 1 5}{1 0} = 0. 1 1 5 \mathrm {k m / s e c} = 0. 1 1 5 \times 3 6 0 0 = 4 1 4 \mathrm {k m / h r}$$

Answer: The speed of the plane is $414\mathrm{km / hr}$