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 3000 ft over ground with an increasing slope of 2°, 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

The problem forms a right triangle with altitude (opposite to 60°) = 3000 ft, CB is adjacent to 60°.

Step2: Use tangent function

$\tan(60^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{3000}{CB}$

Step3: Solve for CB

$CB = \frac{3000}{\tan(60^\circ)}$, $\tan(60^\circ) = \sqrt{3} \approx 1.732$

Step4: Calculate CB

$CB \approx \frac{3000}{1.732} \approx 1732$

Answer:

1700 (rounded to nearest hundred)