Lesson Video: Using Properties of the 30-60-90 Triangle Mathematics

Learn the exact sine, cosine and tangent ratios for 30 and 60° angles. Apply these exact ratio values to calculate lengths of sides in a range of questions involving right-angled triangles with 30 or 60° angles given.

15:58

Video Transcript

In this video, we are going to introduce a particular triangle with angles of 30 degrees, 60 degrees, and 90 degrees. We’ll see how the trigonometric ratios of sine, cosine, and tangent can be worked out exactly for the 30 degrees and 60-degree angle. We’ll see them in the form of surds. And then, we’ll see how to apply them to working out some lengths in such a triangle.

So, we’re going to start by working out what the sine, cosine, and tangent ratios are for the 30-degree and 60-degree angles. Now, I’m beginning with an equilateral triangle in which all three of the sides are one unit. Obviously, because it’s equilateral, all of the angles are 60 degrees. Now, in order to work out the sine, cosine, and tangent ratios, we need right-angled triangles, and that isn’t what I have at the moment. But I can create a right-angled triangle by dividing this triangle in half.

If I draw in the perpendicular height from the top vertex down to the centre of the opposite side, then this will divide the equilateral triangle into two right-angled triangles. Now, just focus on one of these two triangles. And thinking about the angles, I obviously have the 90-degree angle for the right angle. Then, this angle here is still 60 degrees because that’s still the angle from the equilateral triangle I started with. The remaining angle in the triangle, well, I cut that angle in half when I drew in this perpendicular height, so that is half of the 60-degree angle. So, that’s the 30-degree angle. So, this triangle is the 30-60-90-degree triangle that I’m interested in.

We can also work out the lengths of the sides in this triangle. Now, I already know that the hypotenuse is one unit. And the base of this triangle here, well, it was one unit for the full triangle. Now, that I’ve divided it in half, this base here will be a half. Finally, I’d like to work out the third side of this triangle, which I’ve referred to as 𝑥. And I can do that using the Pythagorean theorem.

The Pythagorean theorem, remember, tells you that in a right-angled triangle, if you take the two shorter sides and square them and add them together, you get the same result as if you square the hypotenuse. So, for this triangle, this means if I do 𝑥 squared and a half squared and add them together, I get the same result as if I square one. So, this gives me an equation that I can solve in order to work out the length of this side 𝑥.

Now, I’ll replace a half squared and one squared with their values. So, I have that 𝑥 squared plus a quarter is equal to one. Subtracting a quarter from both sides, I then have that 𝑥 squared is equal to three-quarters. And then, square rooting tells me that 𝑥 is equal to the square root of three over four, or that is equivalent to just the square root of three over two. So, I now have the lengths of all three sides in this right-angled triangle.

Now, I’ve just erased the other half of the original equilateral triangle so that you can see it more clearly. Now, I’m gonna start off with the sine, cosine, and tangent ratios for this angle of 60 degrees. So, I’m gonna label the three sides of this right-angled triangle in relation to that angle of 60 degrees. So, I need the opposite, the adjacent, and the hypotenuse. Now, I’m gonna work out the values of these three ratios for the angle of 60 degrees, and I’m gonna use SOHCAHTOA to help me.

So, sin, first of all, SOH tells me that sine is equal to the opposite divided by the hypotenuse. So, for 60 degrees, looking at the triangle, that’s root three over two divided by one. Now, of course, dividing by one has no effect. Therefore, this just simplifies to sine of 60 is equal to root three over two. Cos, secondly, well, SOHCAHTOA tells me that the cos ratio is the adjacent divided by the hypotenuse. So, looking at the triangle, that is a half divided by one. And again, dividing by one has no effect, so it just simplifies to a half.

Finally, I want to do the tan ratio, so TOA tells me that it’s opposite divided by adjacent. And looking at the triangle, I can see that that’s root three over two divided by a half. Now, dividing by a half is equivalent to multiplying by two. So, that will have the effect of cancelling out the two in the denominator of that first fraction, leaving me with just root three overall. So, those are the exact values of the trigonometric ratios for an angle of 60 degrees. We have them exactly in terms of these surds involving root three.

You can, of course, evaluate them as decimals. But as soon as you do that, you will lose some accuracy in your answer because you’d have to round it to a certain number of decimal places. Consequently, whenever we’re working with an angle of 60 degrees, we use these exact values instead.

Right, now, let’s do the same thing for the angle of 30 degrees. So, I’ll delete the labels for the sides in relation to the 60-degree angle and relabel them in relation to the 30-degree angle. So, the hypotenuse stays the same because it’s always the same side, but the opposite and the adjacent have swapped around. So, now, I’m gonna write down the three trigonometric ratios for this 30-degree angle, starting with the sine ratio.

So, sine is opposite divided by hypotenuse. And in this, instance it will be a half divided by one, which would just simplify to a half. And that gives me my first trigonometric ratio. For cos then, it will be the adjacent divided by the hypotenuse, so that will be root three over two divided by one which would just simplify to root three over two. So, that gives me the second trigonometric ratio for 30 degrees. Finally, looking at tan, tan is opposite divided by adjacent. So, I’ll have a half divided by root three over two.

Now, I’ll have to do a little bit of careful simplification here. Dividing by a fraction, well, that’s equivalent to inverting the fraction and multiplying. So, this is the same then as a half multiplied by two over root three. Now, the two in the numerator and the two in the denominator cancel out, leaving me with just one over root three. But that leaves me with a surd in the denominator. And so, if I go on and if I rationalize that denominator, then it gives me an equivalent answer of root three over three. So, that gives me the three trigonometric ratios in their exact form for an angle of 30 degrees.

Now, just have a look at these ratios for the two angles and you should see some interesting observations. If you look at the sin of 30 and cos of 60, they’re both equal to a half. Similarly, cos of 30 and sin of 60, they’re both equal to root three over two. So, the sine and cosine ratios for these two angles have swapped around, and that’s because the opposite and the adjacent sides swap in relation to those two angles.

If you look at the tan ratios, well, tan of 60 is root three. And within the working out for 30, we have tan of 30 is one over root three. So, they are the reciprocal of each other. And again, that’s due to the opposite and the adjacent swapping around. So, you need to learn these ratios off by heart and be able to recall them. We’ll now look at a couple of questions where we use these ratios.

In this question, we’re given a diagram of a right-angled triangle and we’re asked to calculate the length of 𝐴𝐵.

Now, we can see from looking at the diagram that the angle we’ve been given is 60 degrees, so this is, therefore, a 30-60-90-degree right-angled triangle and, therefore, one where we’re to be using the exact values of the trigonometric ratios. The first step then is to label the three sides of triangle in relation to that angle of 60 degrees. So, I have the opposite, the adjacent, and the hypotenuse. Now, I can see then that I want to calculate the adjacent and know the hypotenuse, so it’s the cosine ratio I’m going to be using because of the CAH part of SOHCAHTOA.

So, I can write down the cosine ratio for this triangle, replacing the angle 𝜃 with 60 degrees, the adjacent with 𝐴𝐵, and the hypotenuse with its value of 7.8. So, that gives me this stage of working out here. Now, I’m looking to calculate 𝐴𝐵. So, in order to do that, I need to multiply both sides of this equation by 7.8. And in doing so, I have that 𝐴𝐵 is equal to 7.8 multiplied by cos 60.

Now, if you had a calculator to help you, you could just of course type this directly into a calculator. But 60 degrees is one of those special angles where we need to know the sine, cosine, and tangent ratios off by heart. So, you could very easily be asked this question when you don’t have access to a calculator. You need to recall then what the value of cos of 60 is. And its exact value is just a half. So, you can substitute then this value of cos of 60 into our working out for 𝐴𝐵. We have then that 𝐴𝐵 is equal to 7.8 multiplied by a half, and that’s easy enough to do without calculator. It is 3.9.

So, within this question then, we started off by labelling the three sides in relation to the angle of 60 degrees. We identified the need for the cosine ratio. Within our working out, we used cosine of 60. And we had to recall that exact value off by heart without using a calculator to give us that value.

In this question, we’re given a right-angled triangle and we’re asked to find the length of the side 𝐴𝐶.

Now, looking at the diagram, we can see that one of the angles marked is 30 degrees. And therefore, we have the 30-60-90-degree triangle. So, we’re going to be using exact values within this question. So, the first step then is to label the three sides of this triangle in relation to the angle of 30 degrees. So, I have the opposite, the adjacent, and the hypotenuse.

Now, looking at the diagram, I’ve been given the opposite and I want to calculate 𝐴𝐶, which is the hypotenuse. So, that tells me I’m going to be using the sine ratio because O and H appear together in the SOH part of SOHCAHTOA. So, I’m gonna write down the sine ratio using the information in this question. I’m going to replace 𝜃 with its value of 30 degrees, the opposite with 7.5, and the hypotenuse, which is what I’m looking to calculate, with 𝐴𝐶.

So, I now have an equation that I’m looking to solve in order to calculate 𝐴𝐶. As 𝐴𝐶 appears in the denominator, the first step is gonna be to multiply both sides of this equation by 𝐴𝐶. And doing so gives me 𝐴𝐶 sin 30 is equal to 7.5. Next, to get 𝐴𝐶 on its own, I need to divide both sides of this equation by sin 30. So, I have 𝐴𝐶 is equal to 7.5 over sin 30.

Now, this is where the fact that it’s a 30-60-90-degree triangle is important. Suppose I haven’t got a calculator for this question. So, rather than using my calculator to tell me what sin 30 is, I need to recall the exact value for myself. And what you should remember is that sin 30 is equal to a half. So, I can use that value directly within the question then.

It becomes then that 𝐴𝐶 is equal to 7.5 divided by a half. Now, dividing by a half is the same as multiplying by two. And therefore, I have that 𝐴𝐶 is equal to 7.5 multiplied by two which is 15. So, within this question, we identified the need for the sine ratio because it was the opposite and hypotenuse still involved. We then had to recall the exact value of sin 30 as being a half because 30 degrees is one of those special angles for which we need to know the values of sine, cosine, and tangent.

In this question, we’re given a diagram of a right-angled triangle and we’re asked to find the length of 𝐵𝐶, giving our answer to two decimal places.

So, first of all I’m gonna label the three sides of this triangle with their names in relation to the angle of 30 degrees. Doing that then tells me that it’s a tan ratio that I’m going to be using here because I’m given the length of the opposite side and I want to work out the length of the adjacent side, so TOA is where opposite and adjacent appear together. So, I recall then the definition of the tangent ratio, which is that tan of an angle is equal to the opposite divided by the adjacent. And now I’m going to write it down for this particular triangle.

I have then that tan of 30 is equal to six over 𝐵𝐶. I want to solve this equation in order to work out the value of 𝐵𝐶, so the first thing I’m going to do is multiply both sides of the equation by 𝐵𝐶 as it currently appears in the denominator of a fraction. I have then that 𝐵𝐶 tan 30 is equal to six. The next step is I need to divide both sides of the equation by tan of 30. I have then that 𝐵𝐶 is equal to six over tan 30.

Now, 30 degrees, remember, is one of those special angles for which we need to recall the exact values of the trigonometric ratios as surds. So, I need to remember what the value of tan 30 is equal to. And if I recall, it’s equal to root three over three, so I’m going to use this value within my calculation. I have then that 𝐵𝐶 is equal to 6 divided by root three over three. Now, I’m dividing by a fraction, so in order to do that I’m going to invert the fraction and multiply. So, this becomes 6 multiplied by three over root three.

This gives me an answer then of 18 over root three. If I didn’t have a calculator, I could leave my answer like this or perhaps I’d actually want to rationalize that denominator, so multiplying by root three over root three. But as this question says, to two decimal places, I do have a calculator, so I can evaluate it. It gives me an answer of 10.39230. And if I round it then to two decimal places, we have then that 𝐵𝐶 is equal to 10.39. And that’s just units because the original question didn’t specify the units for that measurement of six.

So, within this question, we identified the need for the tan ratio because the opposite and adjacent sides were the two sides involved. And then, because the angle was 30 degrees, we had to recall the exact value of tan 30 in terms of a surd and use that value within our calculation.

In summary then, we’ve seen what the exact values of the three trigonometric ratios are for angles of 30 degrees and 60 degrees. We’ve seen where they came from by starting with an equilateral triangle with sides of one unit and dividing it in half in order to create this right-angled triangle. And then, we’ve applied these exact values to a number of problems.

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