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 9°, 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 the right triangle

The altitude (height) = 3200 ft, angle at camera = 60°, CB is opposite angle 30°. Angle at B = 90° - 60° = 30°.

Step2: Use tangent for CB

$\tan(60^\circ) = \frac{3200}{CB}$ → $CB = \frac{3200}{\tan(60^\circ)}$

Step3: Calculate CB

$\tan(60^\circ) = \sqrt{3} \approx 1.732$ → $CB \approx \frac{3200}{1.732} \approx 1847$

Answer:

1800 (rounded to nearest hundred)