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°.
To solve this problem, we can use trigonometry. Let's use the variable x to represent the distance between the lighthouse and the capsized boat, and y to represent the distance between the lighthouse and the patrol boat.
In the triangle formed by the lighthouse, the capsized boat, and the horizontal line, the angle of depression is 12.5°, and the opposite side is x. We can use tangent to relate these values:
tan(12.5°) = x/90
Rearranging the equation, we find:
x = 90 * tan(12.5°)
To find the distance between the lighthouse and the patrol boat, we can use the same logic. In the triangle formed by the lighthouse, the patrol boat, and the horizontal line, the angle of depression is 9°, and the opposite side is y.
tan(9°) = y/90
Rearranging the equation, we find:
y = 90 * tan(9°)
Now we can calculate the values of x and y.
x = 90 * tan(12.5°) ≈ 18.52 m
y = 90 * tan(9°) ≈ 14.69 m
Therefore, the distance between the lighthouse and the capsized boat is approximately 18.52 meters, and the distance between the lighthouse and the patrol boat is approximately 14.69 meters.
To solve this problem, we can use trigonometric ratios involving angles of depression.
Let's label the points involved in this problem:
- The top of the lighthouse is labeled as A
- The capsized boat is labeled as B
- The patrol boat is labeled as C
Given:
- The height of the lighthouse AB = 90 m
- The angle of depression from the top of the lighthouse to the capsized boat, ∠BAD = 12.5°
- The angle of depression from the top of the lighthouse to the patrol boat, ∠CAD = 9°
To find the distances AB and AC, we can use tangent trigonometric ratio:
tan ∠BAD = AB/BD
tan ∠CAD = AC/CD
First, let's find the distance BD (distance from the top of the lighthouse to the capsized boat B). We can use the tangent of ∠BAD:
tan 12.5° = AB/BD
tan 12.5° = 90/BD
BD = 90 / tan 12.5°
Next, let's find the distance CD (distance from the top of the lighthouse to the patrol boat C). We can use the tangent of ∠CAD:
tan 9° = AC/CD
tan 9° = 90/CD
CD = 90 / tan 9°
Now we have the distances BD and CD from the top of the lighthouse to the capsized boat and the patrol boat respectively.