Asked by belle
A passenger on a cruise ship is positioned 42m above sea level and is admiring a tropical island in the distance.
One end of the island is at a bearing of 028° and has an angle of depression of 11°, while the other is at a true
bearing of 118° and has an angle of depression of 19°. How long is the island?
One end of the island is at a bearing of 028° and has an angle of depression of 11°, while the other is at a true
bearing of 118° and has an angle of depression of 19°. How long is the island?
Answers
Answered by
belle
I can't draw it out!! thx for ur help guys!!
Answered by
R_scott
distance to one end ... 42 m / tan(11º)
distance to other end ... 42 m / tan(19º)
118º - 28º = 90º ... the length of the island is the hypotenuse
(length)^2 = (one distance)^2 + (other distance)^2
distance to other end ... 42 m / tan(19º)
118º - 28º = 90º ... the length of the island is the hypotenuse
(length)^2 = (one distance)^2 + (other distance)^2
Answered by
henry2,
d1 + d2. = Hor. distance from cruise to far end of island.
d2 = length of island.
Tan19 = 42/d1,
d1 = 122 m.
Tan11= 42/(d1+d2).
d1+d2 = 216 m.
d2 = 216-d1 = 216-122 = 94 m = Length of island.
d2 = length of island.
Tan19 = 42/d1,
d1 = 122 m.
Tan11= 42/(d1+d2).
d1+d2 = 216 m.
d2 = 216-d1 = 216-122 = 94 m = Length of island.
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