Video: GCSE Mathematics Foundation Tier Pack 4 • Paper 1 • Question 20

GCSE Mathematics Foundation Tier Pack 4 • Paper 1 • Question 20


Video Transcript

Part a) State the exact value of sin of 60 degrees.

Now, sin is one of our three trigonometric ratios: sin, cos, and tan, which describe the ratio that exist between the lengths of the sides of right-angled triangles with particular angles. Normally, when we’re using sin, cos, or tan, we have access to a calculator, but here we don’t. And we’ve been asked to state the exact value of sin of 60 degrees. This is because 60 degrees is a special angle where we can write down the values of sin, cos, and tan exactly in terms of surds or square roots.

There’re other special angles for which we can do the same thing. And those angles are 30 degrees and 45 degrees. There’s a trick to help you remember the values of sin and cos for these three angles. So we’ve written 30 degrees, 45 degrees, and 60 degrees across the top and sin and cos down the vertical side of this table.

And first, what we do is we write one, two, three across the top line for sin and then three, two, one across the bottom line for cos. We then turn each of these values into fractions with a denominator of two. So we have one over two, two over two, and three over two in the top row for sin and three over two, two over two, and one over two in the bottom row for cos.

Next, we include the square root of each of the numerators — only the numerators, not the denominators. Now, the square root of one can actually be simplified as the square root of one is actually just one because one multiplied by one is equal to one. So sin of 30 degrees and cos of 60 degrees can just be simplified to one over two.

Now, that’s it! By remembering that method, we can complete the table of the exact values of sin and cos for 30 degrees, 45 degrees, and 60 degrees. We just need to write down the value that we were asked for, which was sin of 60 degrees. Sin of 60 degrees is equal to root three over two.

Part b) Using the diagram below, work out the value of 𝑥.

Now, the diagram we’ve been given is a diagram of a right-angled triangle, where one of the other angles is 60 degrees. So this suggests that we’re going to be using what we’ve done in part a at some point in this question. In any case, we’re going to need to apply some trigonometry to work out the value of 𝑥 as we’ve got a right-angled triangle, in which we know one side, we know one other angle, and we want to calculate the length of the second side.

For me, the first step with any problem involving trigonometry is to write down the acronym SOHCAHTOA, which tells us which of the trigonometric ratios sin, cos, and tan use which pairs of sides. So S, C, and T stand for sin, cos, and tan and O, A, and H stand for opposite, adjacent, and hypotenuse, which are the sides of this right-angled triangle.

Next, we’ll label the sides of this triangle. So firstly, we have the hypotenuse which is always the longest side and the side opposite the right angle. We also have the opposite which is the side opposite the given angle of 60 degrees and finally the adjacent which is the side between the right angle and the given angle of 60 degrees.

So the side that we know is the adjacent. And the side that we’re looking to work out is the hypotenuse. So A and H appear together in the CAH part of SOHCAHTOA, which tells us that it is the cos ratio we’re going to be using in this question.

The definition of the cos ratio is that cos of an angle 𝜃 is equal to the length of the adjacent side divided by the length of the hypotenuse. In our triangle, the angle 𝜃 is 60 degrees, the adjacent is four centimeters, and the hypotenuse which is what we’re looking to calculate is 𝑥 centimeters. So we have the equation cos of 60 degrees is equal to four over 𝑥.

Now, we want to solve this equation for 𝑥. But first of all, we need to know what the value of cos of 60 degrees is. Well, looking back at the table that we wrote down in part a, we can see that cos of 60 degrees is actually just equal to the fraction one-half. So we can substitute this into our equation. And it gives a half is equal to four over 𝑥.

Now, 𝑥 is currently in the denominator of the fraction on the right of this equation. So our next step in solving is going to be to multiply both sides of the equation by 𝑥. When we multiply the left of the equation by 𝑥, we have 𝑥 multiplied by a half which is equal to 𝑥 over two. And on the right, when we multiply by 𝑥, this will cancel out the 𝑥 in the denominator. So we’re just left with four.

Now, we have 𝑥 over two or 𝑥 divided by two on the left of the equation. So we need to multiply both sides of the equation by two. Multiplying 𝑥 over two by two will just leave us with 𝑥 and multiplying four by two gives eight.

So we’ve answered the problem and found that the value of 𝑥 is eight.

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