Video: Resultant Motion and Force

A surveyor walks through a field, as shown in the diagram. How much further east does the surveyor walk than he walks north? Round your answer to the nearest meter.

05:35

Video Transcript

A surveyor walks through a field as shown in the diagram. How much further east does the surveyor walk than he walks north? Round your answer to the nearest meter.

Okay, so we’ve been told in this question that a surveyor is walking through a field. And the path that the surveyor takes is shown in the diagram. So the surveyor starts here at the origin of the axis that we’ve drawn and finishes here. The total distance that he walks is 450 meters as we’ve been told in the diagram. We also know that he walks at an angle of 30 degrees to east, where east is this way — once again shown to us in the diagram.

What we’ve been asked to do is to find out how much further east does the surveyor walk than he walks north. So what does this mean? Well, we know that he walks in this direction. Now, that direction can be broken up into an eastwards component and a northwards component. Specifically, the eastwards component is this distance here because that’s how much further east this point is relative to this point. And similarly, the northwards component is this distance here because you guessed it that’s how much further north this point is relative to this point.

Now, what the question wants us to do is to work out how much further east the surveyor walks compared to how far he walks north. In other words, how much further is this distance compared with this distance, which means we know we need to find these distances now. So let’s give each one a name. Let’s call this one 𝑥 and this one 𝑦. So how far he walks east is 𝑥 and how far he walks north is 𝑦.

Now, an important thing to know about compass directions east and north is that they’re at right angles to each other. Therefore, this is a right angle. And so we’ve got ourselves a right-angled triangle. The distance that he walks — the 450 meters — and the two components — the eastwards and the northwards components — form the other two sides. So we can draw this triangle a lot more simply. So here’s our right-angled triangle with this being the right angle. Now, we want to work out the distances 𝑥 and 𝑦.

We know one angle and the length of the hypotenuse. So we need to use what we’ve learnt in maths. We need to use SOHCAHTOA. Let’s first try to find the value of 𝑥 using SOHCAHTOA. Well, the angle that we’ll be working with is this one here. Now, relative to that angle, 𝑥 is the adjacent side. And we already know the hypotenuse, which is 450 meters. So in this case, we’re trying to work out the adjacent and we know the hypotenuse. So we need to use cosine.

In other words, we need to use CAH out of SOHCAHTOA. The CAH part tells us that the cosine of an angle 𝜃 is equal to the adjacent 𝐴 divided by the hypotenuse 𝐻. Now, in this case, we already know what 𝜃 is. We know that 𝜃 is 30 degrees. That’s the angle we’re working with. And as well as this, we want to work out what the adjacent is, which we know is 𝑥. So we replace the 𝐴 for adjacent with 𝑥 and we also know the hypotenuse which is 450 meters.

What this means is that we can calculate the left-hand side by plugging in to our calculator. And we know the hypotenuse. So we can work out what 𝑥 is. To do this though, we need to rearrange the equation. What we can do is to multiply both sides of the equation by 450, meaning the 450 is on the right-hand side cancel. And so what we’re left with is that 450 times cosine of 30 degrees is equal to 𝑥. We can then plug this in to our calculator to give us 𝑥 is equal to 389.7114 dot dot dot meters.

So let’s write that down on the bottom right-hand side of the screen and let’s now try and work out what the value of 𝑦 is. Now relative to the angle 30 degrees, 𝑦 is the opposite side. And once again, we already know the hypotenuse. In other words, we’re trying to calculate what the opposite is and we know the hypotenuse. So we need to use sine.

We need to use SOH from SOHCAHTOA. SOH tells us that the sine of the angle 𝜃 — once again this 𝜃 is going to be 30 degrees — is equal to the opposite length divided by the hypotenuse. And so we can do a similar thing to what we did earlier. Sine of 30 degrees where 30 degrees is 𝜃 becomes the opposite which is 𝑦 divided by the hypotenuse which is 450 meters. And yet again, we can rearrange by multiplying both sides of the equation by 450 so that 450 cancels on the right-hand side, leaving us with 450 times sine of 30 degrees is equal to 𝑦. We can plug this into our calculator to give us 𝑦 is equal to 225 meters. So we write that down on the bottom right.

Now what we want to do in this question is to work out how much further east the surveyor walks compared to how far he walks north. In other words, how much larger is 𝑥 compared to 𝑦? And the way to do this is to work out what the value of 𝑥 minus 𝑦 is because 𝑥 minus 𝑦 is how much larger 𝑥 is relative to 𝑦. Now happily, we already know what 𝑥 is and we know what 𝑦 is. So we can plug in the values. 389.7114 dot dot dot is 𝑥 and 225 is 𝑦 and this evaluates to 164.7114 dot dot dot meters.

So is that our final answer? Well, no, we’ve been told to round our answer to the nearest meter. In order to do this, we need to round the last value before the decimal point. That’s the four. Now, it’s the number after that that will tell us whether the value rounds up or stays the same. So in this case, we’ve got a seven and seven is larger than five. So this value will round up. This will now become a five.

And this leads us to our final answer. 𝑥 minus 𝑦 — in other words the amount of distance the surveyor walks further east compared to how far he walks north — is equal to 165 meters to the nearest meter.

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