# Hello, could use some help

Hi, I'm a sophomore in high school at the moment so I don't know much math, however it is my favorite subject..but anyways today we were doing reviews for our EOC (end of course) test for in my geometry class and I came across a funny problem. I want to ask you if this problem makes sense the way it was asked, because I couldn't figure it out as it seemed illogical. Anyways, the problem reads exactly,

"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

$$\sqrt{a^{2}\oplus b^{2}}= c$$

you end up with

$$\sqrt{52}=7.1$$

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

So, to find the angle, you would do

$$\sin{-1}\cdot (4\div7.1)$$

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.

If your teacher told you to use 6 as the hypotenuse and 4 as one of the legs, he or she is wrong. The way you solved it later on is correct, setting 6 and 4 as the two legs and finding the hypotenuse so that you can use inverse sine. Alternatively, you could also just do:

$$\theta = \tan^{-1}\left(\frac{4}{6}\right)$$​

The important part is that you are solving for the right angle.

There is nothing wrong about how the question is worded. 6 mph down and 4 mph to the right, for example, are velocities, not speeds. The distinction is made clear if you take a physics course, but the idea is that velocity is speed with direction. You can add velocities, but you need to take direction into account, hence your need to use Pythagoras's theorem and a right triangle. (Quantities like velocity with both a numeric value and a direction are called vectors, if you want to look them up).

For right now, you can do exactly what you are doing, setting an arbitrary time of 1 hour, finding the distance, and getting the angle that way. Notice, however, that no matter what time period you choose, you will still find the same angle, because the triangles will always be similar. That is because what really determines the angle is in what direction you move, not how far in that direction you move. The velocity of the boat tells you not only how fast you move but also the direction. When you add the velocity vectors from the current and from the boat's motor, you get the total velocity of the boat, which tells you the direction in which the boat is moving.

You represent a vector with an arrow pointing in the direction of the vector, with its length being the numeric value of the vector (called its magnitude or size). The arrow end of the vector is the head, the other is the tail. To add two vectors, you place the tail of one vector on the head of the other. The sum is the vector connecting the tail of the first to the head of the second, and its magnitude is determined, if the vectors are at right angles to each other, by the Pythagorean theorem.

By the way, to display a plus sign, you just need to use +. LaTeX understands that.

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Thank you for your quick response.

I remember a while back I was reading a physics book on vectors (which I had forgotten until you re-explained it to me here) and it had a similar boat problem, and explained how to solve it.

Now I understand the original problem, thank you very much for your help.