To solve this problem, we can use trigonometry. Let's assume that the distance from the lighthouse to the capsized boat is x and the distance from the lighthouse to the patrol boat is y.
Using the angle of depression of 12.5 degrees to the capsized boat, we can set up the trigonometric equation:
tan(12.5°) = x / 90
Rearranging the equation, we get:
x = 90 * tan(12.5°)
Calculating this value, we find:
x ≈ 22.92 m
So, the distance from the lighthouse to the capsized boat is approximately 22.92 m.
Similarly, using the angle of depression of 9 degrees to the patrol boat, we can set up the trigonometric equation:
tan(9°) = y / 90
Rearranging the equation, we get:
y = 90 * tan(9°)
Calculating this value, we find:
y ≈ 14.98 m
So, the distance from the lighthouse to the patrol boat is approximately 14.98 m.
To find the distance between the two boats, we can use the concept of the distance formula. The distance between the two boats is the difference between their distances from the lighthouse:
distance = |x - y|
Substituting the given values, we get:
distance = |22.92 - 14.98|
Calculating this value, we find:
distance ≈ 7.94 m
So, if the two boats are on the opposite side of the lighthouse, they are approximately 7.94 meters apart.
From the top of a 90 m lighthouse, an operator sees a capsized boat and determines an angle of depression of
12.5° to the boat. A patrol boat is also spotted at an angle of depression of 9°.
How far from the lighthouse is the patrol boat?
How far from the lighthouse is the capsized boat?
If the two boats are on the opposite side of the lighthouse, how far apart are the two boats?
1 answer