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)

The angle of depression decreases (from 67° to 22°), meaning the car is moving away from the building. A sketch would show a vertical building (height 38m), horizontal ground, two horizontal lines from the top (parallel to ground), two lines of sight (depression angles) to the car at two positions: closer initially (larger angle, since angle of depression is related to adjacent side in right triangle; smaller adjacent side means larger angle) and farther later (smaller angle, larger adjacent side).

Step1: Recall tangent formula

In a right triangle, $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$. The opposite side is the building height ($h = 38$ m). Let $d_1$ be the initial distance, $d_2$ be the later distance.

Step2: Calculate initial distance $d_1$

For $\theta_1 = 67^\circ$, $\tan(67^\circ)=\frac{38}{d_1}\implies d_1=\frac{38}{\tan(67^\circ)}$.
$\tan(67^\circ)\approx2.3559$, so $d_1\approx\frac{38}{2.3559}\approx16.13$ m.

Step3: Calculate later distance $d_2$

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

Step4: Find distance traveled

Distance $= d_2 - d_1\approx94.06 - 16.13\approx77.93\approx78$ m.

Step1: Convert distance to km and time to hours

Distance traveled: $78$ m $= 0.078$ km. Time: $5$ seconds $=\frac{5}{3600}$ hours.

Step2: Calculate speed

Speed $=\frac{\text{distance}}{\text{time}}=\frac{0.078}{\frac{5}{3600}}=\frac{0.078\times3600}{5}=56.16$ km/h.

Step3: Compare with speed limit

Speed limit is 50 km/h. $56.16>50$.

Answer:

It is moving away from the building. (Sketch: Vertical line for building (height 38m), horizontal line at top. Two right triangles: first with angle of depression 67° (shorter horizontal leg, closer car), second with 22° (longer horizontal leg, farther car), car moves from first to second position along ground.)

Part (b)