An obsever is looking out to sea from the top of a building that is 30 m above sea leavel. He observera ship at an angle of depression of 10. How far is the ship from the foot of the bulding?
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Let O represent the observer,
B = base of the building (vertically below the observer O), and
S = the observed ship.
OBS represents a right-triangle right-angled at B.
Angle of depression
=∠BSO
= 10°
Solve the right triangle:
OB = 30m
&BSO = 10°
tan(10°)=OB/BS = 30m / BS
Solve for BS
BS = 30m /tan(10°)
=30m / 0.17633 (approx.)
= 170 m (approx.)
To find the distance between the ship and the foot of the building, we can use trigonometry. Specifically, we can use the tangent function since we have the angle of depression and the height of the building.
Let's define some variables:
- Let "x" represent the distance between the ship and the foot of the building.
- Let "θ" represent the angle of depression, which is 10 degrees in this case.
We know that the tangent of an angle is equal to the opposite side divided by the adjacent side. In this scenario, the opposite side is the height of the building (30 m), and the adjacent side is the distance between the ship and the foot of the building (x). So, we have the equation:
tan(θ) = opposite/adjacent
tan(10) = 30/x
Now, we can solve for "x" using algebraic manipulation. First, isolate "x" by multiplying both sides of the equation by "x":
x * tan(10) = 30
Next, divide both sides of the equation by tan(10):
x = 30 / tan(10)
Using a calculator, we can evaluate the expression to find:
x ≈ 162.79 m
Therefore, the ship is approximately 162.79 meters away from the foot of the building.