A submarine dives from the water surface at an angle of 31° below the horizontal, following a straight path 75 m long. How far is the submarine then below the water surface?

To find the distance the submarine is below the water surface, we can use basic trigonometry.

1. Draw a diagram to visualize the problem. The submarine starts at the water surface and dives downwards at an angle of 31° below the horizontal.

2. In the diagram, label the distance the submarine travels along the horizontal axis as "x" and the distance below the water surface as "h" (the unknown we are trying to find).

3. Apply trigonometry to the right triangle formed by the submarine's path. The angle between the horizontal distance traveled (x) and the hypotenuse (75 m) is 31°.

4. Recall that the trigonometric function tangent (tan) relates the opposite side (h) to the adjacent side (x). In this case, we have tan(31°) = h/x.

5. Rearranging the equation, we have x = h/tan(31°).

6. Substitute the known values: x = 75 m and tan(31°) = 0.6009.

7. Solve for h using the equation x = h/tan(31°). Rearranging again, we have h = x * tan(31°).

8. Plug in the values: h = 75 m * 0.6009.

9. Calculate the result: h ≈ 45.07 m.

Therefore, the submarine is approximately 45.07 meters below the water surface.

75 sin 31 which is about 38 m because

sin 30 = 1/2