A submarine must descend 516 meters to explore a shipwreck. The submarine begins a constant descent to the shipwreck so that it will reach the shipwreck after traveling 2976 meters from the pint of its initial descent.
What angle of descent did the submarine make?
What horizontal distance did the submarine travel during its descent?
To find the angle of descent, we can use trigonometry. Specifically, we can use the tangent function. Let's call the angle of descent θ.
tan(θ) = Opposite / Adjacent
In this case, the opposite side is the depth the submarine descends, which is 516 meters. The adjacent side is the horizontal distance the submarine travels, which is what we need to find. Let's call it x.
tan(θ) = 516 / x
To find the value of θ, we can take the inverse tangent (arctan) of both sides:
θ = arctan(516 / x)
Now, let's find the horizontal distance the submarine traveled during its descent.
We know that the submarine reaches the shipwreck after traveling 2976 meters from the point of its initial descent. This distance is equal to the hypotenuse of a right triangle formed by the depth and horizontal distance.
Using the Pythagorean theorem, we have:
hypotenuse^2 = depth^2 + horizontal distance^2
(2976)^2 = (516)^2 + x^2
Simplifying the equation, we get:
x^2 = (2976)^2 - (516)^2
x^2 = 8839296 - 266256
x^2 = 8573040
Taking the square root of both sides, we find:
x = sqrt(8573040)
x ≈ 2931.97 meters
Therefore, the horizontal distance the submarine traveled during its descent is approximately 2931.97 meters.
To find the angle of descent, we substitute the value of x into the equation we previously derived:
θ = arctan(516 / 2931.97)
Using a calculator, we find:
θ ≈ 9.97 degrees
Therefore, the angle of descent of the submarine is approximately 9.97 degrees.