Question

A scuba diver has a taut rope connecting the dive boat to an anchor on the ocean floor. The rope is 140 feet long and the water is 40 feet deep. To the nearest tenth of a foot, how far is the anchor from a point directly below the boat?

Answer 1

This can be solved with the Pythagorean Theorem:

h^2=x^2+y^2 where h is the hypotenuse of a right triangle with sides of lenght x and y so:

140^2=40^2+d^2, where we let d=distance from spot below the boat...

d^2=140^2-40^2

d^2=18000

d=√18000

d≈134.16 ft

d≈134.2 ft (to nearest tenth of a foot)

Answer 2

You will need to use the Pythagorean Theorem and plug in the varibles. The rope is the hypotenuse of the right triangle formed, and the 40ft water is the vertical leg, which leaves the ocean floor to be the horizontal leg.[a^{2}+b^{2}=c^{2}\] \[a^{2}+40^{2}=140^{2}\] Then solve for a. \[a^{2}=18000\] \[a=134.164\]