# Chapter 7 - Applications of Trigonometry and Vectors - Section 7.1 Oblique Triangles and the Law of Sines - 7.1 Exercises - Page 304: 46

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.

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