Answer:
[tex]Ship\ A = 40\ miles[/tex]
[tex]Ship\ B = 30\ miles[/tex]
Step-by-step explanation:
Given
[tex]d = 50[/tex] --- distance apart
[tex]Ship\ B = x[/tex]
[tex]Ship\ A = x + 10[/tex]
The question can be represented with the attached image.
Required
Solve for x
Using Pythagoras theorem, we have:
[tex]x^2 + (x + 10)^2 = 50^2[/tex]
Open bracket
[tex]x^2 + x^2 + 20x + 100 = 2500[/tex]
Collect like terms
[tex]x^2 + x^2 + 20x + 100 - 2500 = 0[/tex]
[tex]2x^2 + 20x -2400 = 0[/tex]
Divide through by 2
[tex]x^2 + 10x -1200 = 0[/tex]
Expand
[tex]x^2 + 40x - 30x -1200 = 0[/tex]
Factorize
[tex]x(x + 40) - 30(x +40) = 0[/tex]
Factor out x + 40
[tex](x - 30)(x +40) = 0[/tex]
Split and solve for x
[tex]x - 30 = 0\ or\ x + 40 = 0[/tex]
[tex]x = 30\ or\ x = -40[/tex]
Distance can not be negative
So:
[tex]x = 30[/tex]
Recall that:
[tex]Ship\ B = x[/tex]
[tex]Ship\ A = x + 10[/tex]
This implies that:
[tex]Ship\ B = 30\ miles[/tex]
[tex]Ship\ A = 30 + 10\ miles[/tex]
[tex]Ship\ A = 40\ miles[/tex]
