Question

the distance covered by an aerial photograph is determined by both the focal length of the camera and the tilt of the camera from the perpendicular to the ground. a camera lens with a 6 - in focal length has an angular coverage of ( 86^{circ} ). suppose an aerial photograph is taken vertically with no tilt at an altitude of 2700 ft over ground with an increasing slope of ( 2^{circ} ), as shown in the figure. calculate the ground distance cb that would appear in the resulting photograph. a ground distance of ( square ) ft would appear in the photograph. (round to the nearest hundred as needed.)

Explanation:

Step1: Identify triangle angles

The triangle has angles: 86°, 2°, so the third angle is $180^\circ - 86^\circ - 2^\circ = 92^\circ$. The altitude (opposite 86°) is 2700 ft, and CB is opposite 2°.

Step2: Apply Law of Sines

$$\frac{CB}{\sin(2^\circ)} = \frac{2700}{\sin(86^\circ)}$$

Step3: Solve for CB

$$CB = \frac{2700 \cdot \sin(2^\circ)}{\sin(86^\circ)}$$
$\sin(2^\circ) \approx 0.03490$, $\sin(86^\circ) \approx 0.99756$
$$CB \approx \frac{2700 \cdot 0.03490}{0.99756} \approx \frac{94.23}{0.99756} \approx 94.46$$

Step4: Round to nearest hundred

Nearest hundred of 94.46 is 100.

Answer:

100