A submarine is traveling parallel to the surface of the water of the water 492 feet below the surface. The submarine begins a constant ascent to the surface so that it will emerge on the surface after traveling 3974 feet from the point of its initial ascent.
What angle of ascent did the submarine make?
What horizontal distance did the submarine travel during its ascent?
To find the angle of ascent, we can use the tangent function.
Let's call the angle of ascent θ.
The tangent of an angle is equal to the opposite side divided by the adjacent side. In this case, the opposite side is the vertical distance the submarine traveled (492 feet) and the adjacent side is the horizontal distance it traveled (x feet).
So, we have the equation:
tan(θ) = opposite/adjacent
tan(θ) = 492/x
To solve for θ, we need to find the inverse tangent (also known as arctan or atan) of both sides of the equation:
θ = arctan(492/x)
To find the horizontal distance (x), we can use the Pythagorean theorem.
The Pythagorean theorem states that in a right triangle, the sum of the squares of the two legs (sides adjacent to the right angle) is equal to the square of the hypotenuse (the side opposite the right angle). In this case, the legs are the horizontal distance (x) and the vertical distance (3974 feet), and the hypotenuse is the direct distance the submarine traveled.
So, we have the equation:
x^2 + 3974^2 = (3974 + 492)^2
To solve for x, we can rearrange the equation:
x^2 = (3974 + 492)^2 - 3974^2
x^2 = 4466^2 - 3974^2
x^2 = 19952116 - 15796476
x^2 = 4155640
x ≈ √4155640
x ≈ 2037.78 feet
Therefore, the angle of ascent (θ) is given by θ = arctan(492/2037.78).
We can use a calculator to find the arctan of 492/2037.78, which is approximately 13.85 degrees (rounded to two decimal places).
Hence, the submarine made an angle of ascent of approximately 13.85 degrees.
The horizontal distance the submarine traveled during its ascent is approximately 2037.78 feet.
To find the angle of ascent, we can use trigonometry. We know that the submarine traveled 492 feet downward and 3974 feet horizontally during its ascent.
If we consider a right triangle, with the vertical leg representing the depth of the water (492 ft) and the hypotenuse representing the diagonal distance traveled during the ascent (3974 ft), we can find the angle of ascent using the inverse trigonometric function.
Let's denote the angle of ascent as θ.
Using the tangent function:
tan(θ) = opposite/adjacent = depth/horizontal distance
tan(θ) = 492/3974
Now, we can find the angle θ by taking the arctan (inverse tangent) of both sides:
θ = arctan(492/3974) ≈ 7.0°
Therefore, the angle of ascent is approximately 7.0 degrees.
To find the horizontal distance traveled during the ascent, we can use the cosine function:
cos(θ) = adjacent/hypotenuse = horizontal distance/3974
horizontal distance = cos(θ) * 3974
Substituting the angle of ascent θ ≈ 7.0° into the equation:
horizontal distance ≈ cos(7.0°) * 3974
horizontal distance ≈ 3974 * cos(7.0°)
horizontal distance ≈ 3974 * 0.991446
Therefore, the submarine traveled approximately 3938 feet horizontally during its ascent.