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 3600 ft over ground with an increasing slope of 5°, 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 and angles

The triangle is right-angled (altitude is perpendicular to ground). The angle at the camera is 60°, and the altitude (opposite side to 60°) is 3600 ft. CB is the adjacent side to 60°.

Step2: Use tangent function

$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \implies \tan(60^\circ) = \frac{3600}{CB}$

Step3: Solve for CB

$CB = \frac{3600}{\tan(60^\circ)} = \frac{3600}{\sqrt{3}} \approx 2078.46$

Step4: Round to nearest hundred

2078.46 ≈ 2100

Answer:

2100