The angle of elevation from a ship in the ocean to the top of a lighthouse is 21deg. If the lighthouse is 200 ft. tall, how far out is the boat? Be sure to show work for partial credit.
H = Height of the lighthouse = 200 ft
L = Distance between the lighthouse and a ship
theta = The angle of elevation = 21 °
tan theta = H / L
tan 21 ° = 200 / L Multiply both sides b L
L tan 21 ° = 200 Divide both sides by tan 21 °
L = 200 / tan 21 °
L = 200 / 0.383864
L = 521,01786 ft
To find the distance from the boat to the lighthouse, we can use trigonometry. Specifically, we will use the tangent function.
Let's use the following variables:
- h: height of the lighthouse (200 ft)
- θ: angle of elevation (21 degrees)
- x: distance from the boat to the lighthouse (what we are trying to find)
The tangent of an angle is equal to the ratio of the length of the opposite side to the length of the adjacent side.
In this case, the height of the lighthouse is the opposite side and the distance from the boat to the lighthouse is the adjacent side of the angle of elevation.
Therefore, we can set up the equation:
tan(θ) = h / x
Substituting the known values:
tan(21 degrees) = 200 ft / x
To solve for x, we need to isolate it. We can do this by multiplying both sides of the equation by x:
x * tan(21 degrees) = 200 ft
Now, we can solve for x by dividing both sides of the equation by tan(21 degrees):
x = 200 ft / tan(21 degrees)
Using a calculator, we can find that tan(21 degrees) ≈ 0.3894. Now we can substitute this value and calculate x:
x = 200 ft / 0.3894
x ≈ 514.1343 ft
Therefore, the boat is approximately 514.1343 ft away from the lighthouse.