Video: AQA GCSE Mathematics Higher Tier Pack 1 โ€ข Paper 2 โ€ข Question 24

AQA GCSE Mathematics Higher Tier Pack 1 โ€ข Paper 2 โ€ข Question 24

04:54

Video Transcript

The diagram shows a roundabout in a playground. Johnny is sitting at point ๐ต, and his mum is standing behind a fence at point ๐ด. The fence is perpendicular to the line ๐ด๐ต. Johnny moves anticlockwise to point ๐ถ, which is a perpendicular distance of ๐ฟ from the fence. Angle ๐ถ๐‘‚๐ท is equal to ๐‘ฅ degrees.

Part a) Show that ๐ฟ is equal to six plus three sin of ๐‘ฅ degrees metres.

Itโ€™s not instantly obvious how to answer this problem, but we can break it down into smaller steps. The first step is to find the horizontal distance of the point ๐ท from the fence. Then we find the distance between the points ๐ท and ๐ถ. And adding those two values will give us the length ๐ฟ.

Now we know that the line joining the centre of the circle to its circumference is the radius. So the radius of our circle or our roundabout is three metres. And therefore, we can add in another radius in this circle, which must also be three metres in length. We can see that the roundabout is three metres away from the fence. So the horizontal distance of point ๐ท to the fence can be found by adding three metres and three metres. Thatโ€™s six metres.

Now we just need to find the length of the line ๐ท๐ถ. We can see that the line ๐ท๐ถ makes out part of a right-angled triangle, with an included angle of ๐‘ฅ degrees. The hypotenuse of this triangle โ€” remember, thatโ€™s the longest side and the one that sits opposite the right angle โ€” is the radius of the circle, so it must be three metres. And we can use right angle trigonometry to find an expression for the line ๐ท๐ถ in terms of ๐‘ฅ.

๐‘‚๐ถ we said is the hypotenuse of the triangle, and ๐ท๐ถ is the opposite. Thatโ€™s the side opposite the included angle. The remaining side is the adjacent. Thatโ€™s the one next to the included angle. In fact, we donโ€™t need that side.

We know the length of the hypotenuse, and weโ€™re trying to find the length of the opposite side. We use the sine ratio. Sin ๐œƒ is opposite over hypotenuse. Substituting what we know about our triangle into this formula gives us sin of ๐‘ฅ degrees is equal to ๐ท๐ถ over three.

We can solve this to form an expression for ๐ท๐ถ in terms of ๐‘ฅ by multiplying both sides by three. And that tells us that ๐ท๐ถ is equal to three sin of ๐‘ฅ degrees. We said that the length of ๐ฟ was equal to the distance of ๐ท from the fence, which we said was six metres, plus the distance ๐ท๐ถ. Thatโ€™s six plus three sin ๐‘ฅ. Since our measurements are in metres, ๐ฟ is equal to six plus three sin of ๐‘ฅ degrees metres, as required.

Now at this point, we do need to clear a little bit of space to move on to part b. However, since this is a show that question is hugely important that you leave everything youโ€™ve written down on your paper.

Part b) Johnny continues to spin anticlockwise around the roundabout until he is at point ๐‘ƒ. Work out the length ๐ฟ now.

We can use the formula we just created, this time with an angle of 130 degrees. Substituting 130 degrees into our formula gives us six plus three sin of 130, which is 8.2981 metres.

Now no level of accuracy is specified in the question. However, since this is in metres, if we round to the nearest one hundredth, that will be like rounding to the nearest centimetre. The deciding digit, eight, is greater than five, so that tells us to round the nine up. However, when we round the nine up to a 10, we canโ€™t fit the digits one and zero in this column, so we carry the one and two becomes three. 8.2981 rounds to 8.30 metres. And the length ๐ฟ now is 8.30 metres.

Part c) State the size of angle ๐‘ฅ for which the length ๐ฟ is smallest and state the value of ๐ฟ.

Letโ€™s consider the graph of ๐‘ฆ is equal to sin ๐‘ฅ. Itโ€™s a periodic function; that is, it repeats, and has a period of 360 degrees. Its maximum is when ๐‘ฅ is 90 degrees, and that gives us one. And its minimum is when ๐‘ฅ is equal to 270 degrees. That gives us a value of negative one.

The smallest value in our function then must occur when ๐‘ฅ is equal to 270. We donโ€™t actually need to include the degree symbol here since itโ€™s included in the formula. We know at this point that sin of ๐‘ฅ is equal to negative one. So we can substitute negative one into our formula for ๐ฟ. And that gives us six plus three multiplied by negative one, which is three. The size of angle ๐‘ฅ for which the length ๐ฟ is smallest is 270, and the value of ๐ฟ here is three.

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