Question

a tourist’s horizontal line of sight is 5 ft above ground level.

• the tourist looks up at an angle of elevation of 7° to view the top of a 50-foot-tall building.
• the tourist walks closer to the building, along a straight path, and looks up at an angle of elevation of 27° to view the top of the building again.

which expression represents the distance, in feet, that the tourist walks between the two instances of viewing the top of the building?
a. $\frac{50-5}{sin 7}$
b. $\frac{50-5}{\tan 7}$
c. $\frac{50-5}{sin 7} - \frac{50-5}{sin 27}$
d. $\frac{50-5}{\tan 7} - \frac{50-5}{\tan 27}$

Explanation:

Step1: Determine the vertical height difference

The building is 50 ft tall, and the tourist's line of sight is 5 ft above ground. So the vertical height from the tourist's line of sight to the top of the building is \( 50 - 5 \) feet.

Step2: Recall the tangent function in right triangles

For a right triangle, \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \), where \( \theta \) is the angle of elevation, the opposite side is the vertical height (\( 50 - 5 \)), and the adjacent side is the horizontal distance from the tourist to the building.

Step3: Find the initial and final horizontal distances

Let \( d_1 \) be the initial distance (when angle is \( 7^\circ \)) and \( d_2 \) be the final distance (when angle is \( 27^\circ \)). Using \( \tan(\theta) = \frac{50 - 5}{\text{distance}} \), we solve for distance: \( \text{distance} = \frac{50 - 5}{\tan(\theta)} \). So \( d_1 = \frac{50 - 5}{\tan(7^\circ)} \) and \( d_2 = \frac{50 - 5}{\tan(27^\circ)} \).

Step4: Calculate the distance walked

The distance walked is the difference between the initial and final distances: \( d_1 - d_2 = \frac{50 - 5}{\tan(7^\circ)} - \frac{50 - 5}{\tan(27^\circ)} \).

Answer:

D. \(\frac{50 - 5}{\tan 7} - \frac{50 - 5}{\tan 27}\)