### Video Transcript

A river is one-half mile wide, with a current flowing at two miles per hour from east to west. A man can swim at three miles per hour in still water. He stands on the south bank. So how many degrees east of north should he head at this rate in order to travel directly north across the river? Give your answer correct to one decimal place. What would the answer to this problem be if the river flowed at three miles per hour and the man could swim only at the rate of two miles per hour?

So what we’re gonna do is we’re gonna look at the part of the question with the man swimming at three miles per hour first. So to help us, what we’re gonna do is draw a sketch. So first of all, we have our river. And we know that our river is 0.5 miles wide. Well, next, what we’re gonna look at is the information that’s actually gonna be most useful for this problem. And that is the current of our river and the speed in which the man can swim.

So what I’ve done now is actually drawn two points. One is on the south bank, and that’s where the man started from. And one is on the north bank, and this is directly north. This is point B. And this is where he would like to get to. Well, what we know is that there’s a current flowing at two miles per hour from east to west. And then we know that the man can swim at three miles per hour in still water. Well, what I’ve drawn here is an arbitrary line just to the angle where the man might head off in order to actually reach B at the other side.

Well, what we could see now is that, in fact, what we’ve created is a right-angled triangle, where the angle 𝜃 is what we’re looking to find out, because what we’re trying to find is how many degrees east of north he should head at the rate that we know in order to travel directly across the river, so in order to go from A to B. So we can see that this is the angle that we’re gonna look to find. And we can find this using our trigonometric ratios. So let’s go on and do that.

Well, the first thing we always do when we’re looking to use our trigonometric ratios is label the sides of our triangle. So here we’ve got the hypotenuse, the adjacent, and the opposite. Hypotenuse is our longest side opposite the right angle. And then we’ve got the opposite, which is opposite 𝜃, and the adjacent, which is next to 𝜃. Well, the next stage with any problem like this is to actually choose which ratio to use. Well, we’ve got the opposite because that’s the two miles per hour opposite our 𝜃. And we also know the hypotenuse because this is the longest side.

At this point, there’s one common mistake to be careful to avoid. And that is the fact that we think, oh, we have A; we’ve got the adjacent because it’s 0.5 miles. This is not the case because that is the distance and actually we’re looking at a triangle formed of speeds.

So now what we can do is remind ourselves of a memory aid that we can use to solve this type of problem. And that is SOHCAHTOA. So using that, we can see that we have the O and the H because we have the opposite and the hypotenuse. So therefore, we’re gonna use the sine ratio cause that means that the sine of an angle is equal to the opposite divided by the hypotenuse. So therefore, we can say that sin 𝜃 is gonna be equal to two over three or two-thirds. So therefore, we can say that 𝜃 is going to be the inverse sin or arcsin of two-thirds. Well, this is gonna give us 𝜃 equal to 41.81031 et cetera degrees.

Well, we haven’t quite finished here, because if we check the question, it wants us to leave our answer to one decimal place. So therefore, we can say that if the man stands on the south bank, the degrees east of north that he should head at the rate in order to travel directly north across the river is 41.8 degrees.

Okay, now we’re gonna have a look at the second part of the question, where we’re asked what would happen if the river flowed at three miles per hour and the man could only swim at a rate of two miles per hour. Well, in order to work this out, what we’ve done is changed our diagram or changed our sketch. So we’ve added two miles per hour for where our hypotenuse would be and three miles per hour for where our opposite would be.

However, this doesn’t make sense because the greatest value or the longest distance should be the hypotenuse. So this looks like this is gonna be undefined. And we can quickly show that now in a calculation. Because let’s imagine we hadn’t spotted that straight away and we just carried on with the calculation. So here we’d have sin 𝜃 equals three over two. Well, then we get 𝜃 is equal to the inverse sin of three over two. Well, if we try to calculate that, we could see from our calculator that 𝜃 was in fact undefined.

So therefore, we can see that, for the second part of the problem, it is not possible to head north if the river flows at three miles per hour and the man could only swim at the rate of two miles per hour.