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Math Help - In Terms of - Boating

  1. #1
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    In Terms of - Boating

    A man can row at 24 km/h, and run at 7 km/h. He needs to get from a point A, on the south bank of a stretch of still water, to point B on the north bank of the water. The direct distance from A to B is 25 km, and the water is 24 km wide. He starts rowing with an angle θ between North and the direction in which he rows. Find an expression for the time he will take to get from A to B, in terms of θ.

    I've been stuck on this question for quite sometime now.. would anybody kindly help me out?

    Thank you!
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  2. #2
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    Quote Originally Posted by gearshifter View Post
    A man can row at 24 km/h, and run at 7 km/h. He needs to get from a point A, on the south bank of a stretch of still water, to point B on the north bank of the water. The direct distance from A to B is 25 km, and the water is 24 km wide. He starts rowing with an angle θ between North and the direction in which he rows. Find an expression for the time he will take to get from A to B, in terms of θ.

    I've been stuck on this question for quite sometime now.. would anybody kindly help me out?

    Thank you!
    You are dealing with a right triangle:

    The hypotenuse has the length 25, one leg has the length 24 and therefore the second leg must be 7 (Pythagorean theorem)

    Since you know all three sides of the triangle you can choose the trigonometric function you like. I'll take Cosine function:

    \cos(\theta) = \dfrac{24}{25}~\implies~\theta\approx 16.26^\circ

    \cos(\theta) = \dfrac{24}{25}~\implies~25=\dfrac{24}{\cos(\theta)  }

    The definition of speed is:

    speed=\dfrac{distance}{time}~\implies~time=\dfrac{  distance}{speed}

    Plug in the values you already know:

    t = \dfrac{25}{24} that means you'll get: t = \dfrac{\dfrac{24}{\cos(\theta)}}{24} = \dfrac1{\cos(\theta)} = \sec(\theta)

    I'm a little bit confused about the running speed of the man. You don't need this fact. If - and only if - you are asked to find the minimum time it takes to go from A to B by rowing and running, then, of course, my calculations are wrong.
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