Question

page 187: a weekend challenger...
from the top of a 38 metre tall building, i look
down to the street below. at an angle of
depression of 67 degrees, i see a car driving. five
seconds later, it is now at an angle of depression
of 22 degrees.

a. is it moving towards or away from the building?
show a sketch.

b. how far has it traveled (to the nearest metre)?

c. if the speed limit is 50km per hour, is the car
speeding? explain your conclusion.

Explanation:

Part (a)

Step1: Analyze angle of depression

The angle of depression is equal to the angle of elevation from the car to the top of the building (alternate interior angles). A smaller angle of depression means the car is farther from the building (since \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\), smaller \(\theta\) with same opposite side (building height) means larger adjacent side (distance from building)). So as the angle of depression decreases from \(67^\circ\) to \(22^\circ\), the car is moving away from the building.

Step2: Sketch description

• Draw a vertical line to represent the building (height \(h = 38\) m).
• From the top of the building, draw two horizontal lines (to represent the horizontal line of sight) and two lines of sight making angles of \(67^\circ\) and \(22^\circ\) with the horizontal (angle of depression) to the street.
• Mark the initial position of the car (closer to the building, under the \(67^\circ\) line of sight) and the later position (farther, under the \(22^\circ\) line of sight).

Step1: Find initial distance (\(d_1\)) from building

Using \(\tan(\theta)=\frac{h}{d}\), where \(h = 38\) m, \(\theta_1 = 67^\circ\) (angle of elevation, equal to angle of depression). So \(d_1=\frac{h}{\tan(\theta_1)}\).
\(d_1=\frac{38}{\tan(67^\circ)}\). \(\tan(67^\circ)\approx2.3559\), so \(d_1\approx\frac{38}{2.3559}\approx16.13\) m.

Step2: Find later distance (\(d_2\)) from building

For \(\theta_2 = 22^\circ\), \(d_2=\frac{h}{\tan(\theta_2)}\). \(\tan(22^\circ)\approx0.4040\), so \(d_2\approx\frac{38}{0.4040}\approx94.06\) m.

Step3: Calculate distance traveled

Distance traveled \(= d_2 - d_1\approx94.06 - 16.13 = 77.93\approx78\) m.

Step1: Convert distance and time to speed

Distance traveled \(d = 78\) m, time \(t = 5\) s. Speed in m/s \(v=\frac{d}{t}=\frac{78}{5}=15.6\) m/s.

Step2: Convert speed to km/h

To convert m/s to km/h, use \(1\) m/s \(= 3.6\) km/h. So \(v = 15.6\times3.6 = 56.16\) km/h.

Step3: Compare with speed limit

Speed limit is \(50\) km/h. Since \(56.16>50\), the car is speeding.

Answer:

The car is moving away from the building. (Sketch: Vertical building, two horizontal lines from top, two lines of sight with \(67^\circ\) and \(22^\circ\) angles to horizontal, initial car closer, later car farther.)

Part (b)