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"If a boat is tavelling in still water at 6 miles per hour. It suddenly hits a crosscurrent going 4 miles per hour, at what angle will the boat veer?"

Now, this is why I don't understand the question, what you are supposed to do (as I'm sure you know), is use trigonomic functions to solve the problem. The problem is, there is not a distance in which you can multiply sides by to figure out the angle..right? Well I asked my teacher, I asked her and told her it didn't seem to make sense. She said, "You are supposed to put the number 6 as the hypotenuse, and the number 4 as one of the legs." I said.."I don't see how that makes sense because you can't put a distance if you just have a speed." Because, as you can see, the problem does not give an amount of time. She told me it was a poorly asked question and just to solve it like she said. So I did.

So, besides the fact that this problem seemed to make no sense at all to me, it didn't seem like I would be solving it the correct way had it been a proper question anyways. Or it seems like it to me..

This is how

**I**proposed to solve it. (Try to draw this picture in your head..), I pictured the boat going downward at 6mph as the height of the triangle (it's a right triangle), and the crosscurrent as the base of the triangle going at 4mph. I figured you would solve for the hypotenuse, and the angle from the hypotenuse to the height leg is the angle at which the boat veers. Would this be the correct way to solve the problem (had I been given the distances for the legs of the triangle)?

So, using my solution I revised the problem to read,

"A boat is travelling in still water at 6 miles per hour. It suddenly hits a crosscurrent going 4 miles per hour. At what angle will the boat veer after

**1 hour?**

Would this problem be a 'correct' problem? and, if so, would my answer be right (about to solve it below)?

This is how I would solve it, I picture a triangle, the height is 6 and the base is 4.

Using the Theory of Pythagoras

[tex]\sqrt{a^{2}\oplus b^{2}}= c[/tex]

you end up with

[tex]\sqrt{52}=7.1[/tex]

About 7.1 is the length of the hypotenuse after an hour.

So, to find the angle, you would do

[tex]\sin{-1}\cdot (4\div7.1)[/tex]

Is this correct?

(I don't have the answer because I don't have a calculator nor a graph at the moment and my computer calculator..somehow..is nowhere to be found.)

Sorry if I did the LaTex wrong..don't exactly know how to work it.