Question

from the top of a tower which is 100m height, an observer observes a bus on the ground with an angle of depression 30°,then find the distance between the bus and the tower​

Answer 1

Answer:

173m

Step-by-step explanation:

In Tigonometry

Tan 30° = 1/√3

The distance equals

Tan 30° = 100 / x

( 100 is the height and 'x' is the distance)

1/√3 = 100 / x

x = 100 × √ 3

x = 173

Answer 2

Answer:

Distance between the bus and the tower is 173 meters.

Step-by-step explanation:

Draw a diagram, make sure to mark the angle of depression.

Let AB be the tower = 100m.

Let A be the position of the observer.

Let B be the position of the Bus.

Hence, BC becomes the distance between the bus and the tower.

Angle of depression = 30°

Hence, alternate angle, i.e, the angle of elevation also becomes 30°.

Now, In ΔABC

$\Longrightarrow \sf tan30^\circ = \dfrac{opposite \ side}{adjacent \ side}$

$\Longrightarrow \sf tan30^\circ = \dfrac{100}{BC}$

$\Longrightarrow \sf \dfrac{1}{\sqrt{3}} = \dfrac{100}{BC}$

$\Longrightarrow \sf BC = 100\sqrt{3}$

$\Longrightarrow \sf BC = 100 \times 1.73$

$\Longrightarrow \sf BC = 173m$

Hence, the distance between the bus and the bottom of the tower is 173 meters.

Values used:

√3 = 1.73

tan30° = 1/√3