# Distance Between Two Ships

• MHB
• xyz_1965
In summary, the distance between the two ships is approximately 10,660 feet when the height of the airplane is 3500 feet and the angles of depression to P and Q are 48 ° and 25 ° respectively.

#### xyz_1965

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.

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 °.