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 12 - in focal length has an angular coverage of 60°. suppose an aerial photograph is taken vertically with no tilt at an altitude of 3200 ft over ground with an increasing slope of 8°, as shown in the figure. calculate the ground distance cb that would appear in the resulting photograph. a ground distance of □ ft would appear in the photograph. (round to the nearest hundred as needed.)

Explanation:

Step1: Identify triangle type

Triangle with angles 60° and 8°, right-angled at B? No, use Law of Sines: $\frac{CB}{\sin 60^\circ} = \frac{AB}{\sin 8^\circ}$. Assume AB = 3200 ft (altitude).

Step2: Solve for CB

$CB = \frac{3200 \cdot \sin 60^\circ}{\sin 8^\circ}$
$\sin 60^\circ \approx 0.8660$, $\sin 8^\circ \approx 0.1392$
$CB \approx \frac{3200 \cdot 0.8660}{0.1392} \approx \frac{2771.2}{0.1392} \approx 19908$ ft

Step3: Round to nearest hundred

19908 ≈ 19900 ft

Answer:

19900