A boat heading out to sea starts out at Point AA, at a horizontal distance of 1433 feet from a lighthouse/the shore. From that point, the boat’s crew measures the angle of elevation to the lighthouse’s beacon-light from that point to be 15degrees
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. At some later time, the crew measures the angle of elevation from point BB to be 6degrees
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. Find the distance from point AA to point BB. Round your answer to the nearest tenth of a foot if necessary.
Let's assume that the distance from point AA to point BB is x feet.
We can create a right triangle ABC, where A is the lighthouse, C is point AA, and B is point BB.
Angle CAB is 15 degrees, and angle CBA is 6 degrees.
We can use trigonometric functions to solve for x.
In triangle ABC, we have:
tan(15 degrees) = BC/AC
tan(6 degrees) = BC/(AC + x)
Since BC is the height of the lighthouse, it remains constant and cancels out. We can set up the following equation:
tan(15 degrees) = tan(6 degrees) * (AC)/(AC + x)
Now we can solve for x:
tan(15 degrees) / tan(6 degrees) = AC / (AC + x)
Using the given values, we have:
tan(15 degrees) / tan(6 degrees) = 1433 / (1433 + x)
To solve for x, we can cross-multiply and then divide:
(tan(15 degrees) / tan(6 degrees)) * (1433 + x) = 1433
1433 * tan(15 degrees) / tan(6 degrees) + x = 1433
x = 1433 - 1433 * tan(15 degrees) / tan(6 degrees)
Using a calculator, we find:
x ≈ 106.4 feet
Therefore, the distance from point AA to point BB is approximately 106.4 feet.