Video: AQA GCSE Mathematics Higher Tier Pack 3 β€’ Paper 1 β€’ Question 14

Two boats set off from 𝑋 at the same time. They travel in straight lines such that boat A remains due north of boat B at all times, as shown in the figure. Boat B sails on a bearing 𝐾° greater than boat A. (a) Calculate the distance between points π‘Œ and 𝑍. (b) Which of the following most accurately describes angle π‘Šπ‘π‘‰? [A] Less than 𝐾° [B] Between 𝐾° and 2𝐾° [C] Exactly 𝐾° [D] Greater than 2𝐾°

08:19

Video Transcript

Two boats set off from 𝑋 at the same time. They travel in straight lines such that boat A remains due north of boat B at all times, as shown in the figure. Boat B sails on a bearing 𝐾 degrees greater than boat A.

Part a) Calculate the distance between points π‘Œ and 𝑍. Part b) Which of the following most accurately describes angle π‘Šπ‘π‘‰? The options are less than 𝐾 degrees, between 𝐾 degrees and two 𝐾 degrees, exactly 𝐾 degrees, or greater than two 𝐾 degrees.

So we’ve been given a lot of information in this question. But actually all of it has been summarised for us in the diagram. The fact that boat B sails on a bearing 𝐾 degrees greater than the bearing boat A sails on just means that the angle between the two lines formed by each boat’s journey is 𝐾 degrees. That’s this angle here in the diagram.

Let’s look at what we’ve been asked to do then. Part a says calculate the distance between points π‘Œ and 𝑍. That’s this distance here. And in order to work this out, we need to recognise that there are in fact two similar triangles in this diagram: triangle π‘‹π‘Œπ‘ and triangle π‘‹π‘‰π‘Š.

Now let’s look at why these triangles are similar. This angle of 𝐾 degrees is common to both triangles. They both also have a right angle as one of their internal angles. The third angle in these two triangles will also be the same because we know that the sum of the angles in any triangle is 180 degrees. We found then that these two triangles have three angles in common. And therefore, they’re similar triangles.

How does this help us with calculating the distance between points π‘Œ and 𝑍? Well, π‘Œπ‘ is a side on triangle π‘‹π‘Œπ‘. And in fact, it corresponds with side π‘‰π‘Š on the larger triangle. If we can work out the scale factor for these two similar triangles, we’ll be able to use the length π‘‰π‘Š in order to calculate the length of π‘Œπ‘.

In order to find the scale factor, we need to know the lengths of a pair of corresponding sides on the two triangles. We see that, on the smaller triangle, the length of π‘‹π‘Œ is 90 miles. And if we add the lengths of π‘‹π‘Œ and π‘Œπ‘‰ together β€” that’s 90 plus 120 β€” we see that the length of 𝑋𝑉, which is the corresponding side on the larger triangle, is 210 miles.

To find the scale factor, we can divide the length on the larger triangle by the corresponding length on the smaller triangle, giving 210 over 90. Now this scale factor can be simplified because both the numerator and denominator can be divided by 10, leaving us with 21 over nine. But then both of these numbers can be divided by three, leaving us with seven over three. This tells us then that lengths on the larger triangle are seven over three or two and one-third times as big as the corresponding lengths on the smaller triangle.

Now if we’re going back the other way, so if we want to work out the length of a side on the smaller triangle, this means that we need to take the corresponding length on the larger triangle, which in our case is the length of π‘‰π‘Š. That’s seven miles. And because we’re going from larger to smaller, we need to divide by the scale factor we’ve calculated. So we have that π‘Œπ‘ will be equal to seven divided by seven over three.

We then recall that, in order to divide by a fraction, we flip or invert that fraction and we multiply. So seven divided by seven over three is the same as seven multiplied by three over seven. We can think of the integer seven as seven over one if it helps.

Before we multiply, we can actually cross-cancel. There’s a seven in the numerator of the first fraction and a seven in the denominator of the second. So we’re left with one multiplied by three, which is three, over one multiplied by one, which is one. And three over one is just equal to three. So this tells us that the length of π‘Œπ‘ is three miles.

And we can see straight away that this makes sense. If we recalculate our scale factor using this pair of corresponding sides β€” so that’s π‘‰π‘Š over π‘Œπ‘ β€” we get seven over three, which is equal to the scale factor we’ve already calculated. So we’ve got our answer to part a then. The distance between π‘Œ and 𝑍 is three miles. And now let’s consider part b.

Part b asked us which of the following most accurately describes angle π‘Šπ‘π‘‰. That’s the angle formed when we go from π‘Š to 𝑍 to 𝑉. That’s this angle here, the one marked with the pink question mark. We don’t want to know how big this angle is, just how big it is in relation to the angle of 𝐾 degrees.

Well, let’s start by drawing in this line here, a line which is parallel to the line π‘Œπ‘‰. And so it meets the line π‘‰π‘Š at a right angle. This line divides angle π‘Šπ‘π‘‰ up into two parts, a part labelled with a green dot and a part labelled with an orange dot. So angle π‘Šπ‘π‘‰ will be the sum of these two angles.

The angle marked with a green dot will actually be equal to 𝐾 degrees, because it is a corresponding angle with the angle of 𝐾 degrees in triangle π‘‹π‘Œπ‘. And we know that corresponding angles are equal. So we now know that angle π‘Šπ‘π‘‰ is equal to 𝐾 degrees plus some other angle, which means we can actually rule out two of the multiple choice options we were given straight away. Angle π‘Šπ‘π‘‰ can’t be less than 𝐾 degrees, and it also can’t be exactly 𝐾 degrees.

Now we need to think about this angle marked with an orange dot. And to do so, we need to consider triangle 𝑍𝑉𝑍 prime. To help us visualise what we’re about to do, I’ve drawn this triangle out the other way up. So I’ve flipped it over. The line 𝑍𝑍 prime, which was originally the bottom of this triangle, is now at the top.

We can actually work out some of the lengths in this triangle. The vertical side of this triangle will actually be equal to the side π‘Œπ‘, which we’ve already calculated, because they are two sides which are between the parallel lines π‘Œπ‘‰ and 𝑍𝑍 prime. So we know that the side 𝑍 prime 𝑉 is three miles long. We can also work out the length of the side 𝑍𝑍 prime because it will be the same as the side π‘Œπ‘‰. So the side 𝑍𝑍 prime is 120 miles long.

Now we’re going to compare this triangle with triangle 𝑍𝑍 prime π‘Š. That’s the triangle now marked in orange in the original diagram. We already know that 𝑍𝑍 prime is 120 miles, and we can also work out the length of 𝑍 prime π‘Š. If the total length of π‘‰π‘Š is seven miles and the length of the top portion 𝑉𝑍 prime is three miles, then the length of 𝑍 prime π‘Š will be seven minus three. That’s four miles.

Now if we compare these two triangles, we can see that they have the same base of 120 miles. But we see that the height of the triangle containing the orange angle is less than the height of the triangle containing the green angle, which we already know to be 𝐾 degrees. This tells us that the orange angle must be less than the green angle because the base of the triangles is the same but the height of the triangle with the orange angle is less.

We already know that the green angle is 𝐾 degrees. We’ve shown that already. So we found that the orange angle must be less than 𝐾 degrees. That means, for angle π‘Šπ‘π‘‰, we’re adding 𝐾 degrees to something less than 𝐾 degrees, which means the answer will be less than two 𝐾 degrees. Angle π‘Šπ‘π‘‰ then will be between 𝐾 degrees and two 𝐾 degrees. So we tick the correct option.

We’ve now completed the problem. In part a, we found that the distance between the points π‘Œ and 𝑍 was three miles. And in part b, we found that the angle π‘Šπ‘π‘‰ is most accurately described as being between 𝐾 degrees and two 𝐾 degrees.

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