Question

from the top of a 20 m tall building, a bus is sighted at an angle of depression of 40°. show a sketch of the situation. how far from the building is the bus?

Explanation:

Step1: Understand the problem (angle of depression)

The angle of depression is equal to the angle of elevation from the bus to the top of the building (alternate interior angles). Let the height of the building be \( h = 20\space m \), the distance from the bus to the building be \( d \), and the angle of elevation (equal to angle of depression) be \( \theta=40^\circ \).

Step2: Use the tangent function in trigonometry

In a right - triangle (formed by the building, the ground, and the line of sight from the bus to the top of the building), \( \tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{h}{d} \)

We know that \( h = 20\space m \) and \( \theta = 40^\circ \), so we can re - arrange the formula for \( d \): \( d=\frac{h}{\tan\theta} \)

Step3: Substitute the values

We know that \( \tan(40^\circ)\approx0.8391 \) and \( h = 20\space m \)

\( d=\frac{20}{\tan(40^\circ)}=\frac{20}{0.8391}\approx23.83\space m \)

Answer:

The bus is approximately \( \boldsymbol{23.83\space m} \) far from the building.

(For the sketch: Draw a vertical line segment to represent the building (length 20 m). From the top of this vertical segment, draw a horizontal line. Then draw a line from the top of the building to a point (representing the bus) on the horizontal line (ground level) such that the angle between the horizontal line (from the top of the building) and the line of sight to the bus is \( 40^\circ \) (angle of depression). The horizontal distance between the base of the building and the bus is the distance we calculated.)