- #1
xyz_1965
- 76
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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.
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.