SOLUTION: Two boats leave San Diego at the same time. One boat sails due north and the other due west. When they are 10 miles apart, the boat traveling north has sailed 2 miles farther than

Question 333494: Two boats leave San Diego at the same time. One boat sails due north and the other due west. When they are 10 miles apart, the boat traveling north has sailed 2 miles farther than the boat sailing west. Find the distance each baot has traveled. I am not good at word problems at all! They make me more confused with question.
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
the boats are sailing at a right angle to each other (remember Pythagoras)

when the hypotenuse is 10 mi , one side is 2 mi longer than the other

a^2 + b^2 = c^2

x^2 + (x+2)^2 = 10^2

x^2 + x^2 + 4x + 4 = 100 ___ 2x^2 + 4x - 96 = 0 ___ x^2 + 2x - 48 = 0

factoring ___ (x + 8)(x - 6) = 0

x = -8 ___ and ___ x = 6 ___ negative value NOT realistic

so the shorter distance (west) is 6 mi , and the longer distance (north) is 8 mi

FYI ___ this is a 3-4-5 triangle
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