#### 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.