MHB How Do You Calculate the Distance Between Two Ships Using Angles of Depression?

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To calculate the distance between two ships using angles of depression, the height of the airplane is crucial, set at 3500 feet. The angle of depression to ship P is 48°, leading to a calculated distance PD of approximately 3151.41 feet, while the angle to ship Q is 25°, resulting in a distance DQ of about 7505.77 feet. The total distance between the two ships is the sum of PD and DQ, equating to 10,657.18 feet. After rounding to the nearest tenth, the final distance is 10,660 feet. Understanding the angles of depression is essential for accurate calculations.
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Points P and Q are in the same vertical plane as an airplane at point R. When the height of the airplane is 3500 feet, the angle of depression to P is 48° and that to Q is 25°. Find the distance between the two ships. Round the answer to the nearest 10th of a foot.

Solution:

From R, I will drop a perpendicular to a point I call D. The distance between the two ships is PD + DQ.

To find PD:

tan (48°) = 3500/PD

PD = 3500/tan (48°)

PD = 3151.41 feet

To find DQ:

tan (25°) = 3500/DQ

DQ = 3500/tan (25°)

DQ = 7505.77 feet

Distance between the two ships:

PD + DQ = 3151.41 + 7505.77

PD + DQ = 10, 657.18

Rounded to the neatest 10th of a foot, I get
10,660 feet.

Is this correct? I hope so after all this work.
 
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Are you clear on what "angle of depression" means? Since the angle of depression, the angle down from a horizontal to the line from R to P, is 48 ° the angle RPD is 90- 48= 42 °. Similarly the angle RQD is 90- 25= 65 °.
 
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