Answer
The ground distance that will be shown in this photograph is 5125.9 feet.
Work Step by Step
Let the airplane be at position A. Let the point on the ground directly under the airplane be point D. Then AD = 3500 feet.
Let C be the point on the ground on the far left of the camera's range. Let E be the point on the far right that is on the same horizontal level as point C. Then the triangle ACE is an isosceles triangle. Then angle ACE is $54^{\circ}$.
Let B be the point on the ground on the far right of the camera's range. Then angle $ACB = 49^{\circ}$ and angle $ABC = 59^{\circ}$.
We can use the law of sines to find the distance $CD$:
$\frac{CD}{sin~36^{\circ}} = \frac{3500}{sin~49^{\circ}}$
$CD = \frac{3500~sin~36^{\circ}}{sin~49^{\circ}}$
$CD = 2725.9~feet$
We can use the law of sines to find the distance $BD$:
$\frac{BD}{sin~36^{\circ}} = \frac{3500}{sin~59^{\circ}}$
$BD = \frac{3500~sin~36^{\circ}}{sin~59^{\circ}}$
$BD = 2400.0~feet$
We can find the distance $CB$:
$CB = CD + BD$
$CB = 2725.9+2400.0 = 5125.9~ft$
The ground distance that will be shown in this photograph is 5125.9 feet.
