## Trigonometry (11th Edition) Clone

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 $47^{\circ}$. Let B be the point on the ground on the far right of the camera's range. Then angle $ACB = 42^{\circ}$ and angle $ABC = 52^{\circ}$. We can use the law of sines to find the distance $CD$: $\frac{CD}{sin~43^{\circ}} = \frac{3500}{sin~42^{\circ}}$ $CD = \frac{3500~sin~43^{\circ}}{sin~42^{\circ}}$ $CD = 3567.3~feet$ We can use the law of sines to find the distance $BD$: $\frac{BD}{sin~43^{\circ}} = \frac{3500}{sin~52^{\circ}}$ $BD = \frac{3500~sin~43^{\circ}}{sin~52^{\circ}}$ $BD = 3029.1~feet$ We can find the distance $CB$: $CB = CD + BD$ $CB = 3567.3+3029.1 = 6596.4~ft$ The ground distance that will be shown in this photograph is 6596.4 feet.