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 ( 9^{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 type and angles

The aerial photo forms a right triangle with altitude (opposite side to half the angular coverage) = 2700 ft, half the angular coverage = 43°. Let CB = x (adjacent side).

Step2: Apply tangent function

$\tan(43^\circ) = \frac{2700}{x}$ → $x = \frac{2700}{\tan(43^\circ)}$

Step3: Calculate value

$\tan(43^\circ) \approx 0.9325$ → $x \approx \frac{2700}{0.9325} \approx 2895$

Answer:

2900