### Video Transcript

In this video, we are going to see how the three trigonometric ratios sine, cosine, and tangent can be applied to a mixed set of problems.

So, a reminder then of what the trigonometric ratios are. I have here a diagram of a right-angled triangle, and I’ve labeled one of the other angles as 𝜃. I’ve then labeled the three sides of the right-angled triangle in relation to that angle 𝜃. So, we have the opposite, the adjacent, and the hypotenuse. The trigonometric ratios are the ratios that exist between the different pairs of sides within this triangle.

So, we have their definitions on the screen here. Firstly, sin of 𝜃 is the ratio between the opposite and the hypotenuse, so it’s opposite divided by hypotenuse. Cos of 𝜃, which is usually abbreviated to cos, is the adjacent divided by the hypotenuse. And finally, tan of 𝜃, usually abbreviated to tan, is the opposite divided by the adjacent. So, we’ll now look at a mixed set of problems where these ratios will be needed. And we’ll see how to identify which ratio it is that’s required and then how to apply it.

This problem says, a palm tree 10.6 metres tall is observed from a point 12 metres away on the same horizontal plane as the base of the tree. We’re asked to find the angle of elevation to the top of the palm tree and then give our answer to the nearest minute.

So, to start off with, we haven’t been given a diagram in this question. And I would always suggest that if you’re not given a diagram, then you draw your own. So, we’re going to draw a sketch of the palm tree, the horizontal ground, and then the line of sight from the ground to the top of the palm tree. So, this of course forms a right-angled triangle.

Now, we need to put some information onto this diagram. We’re told the palm tree 10.6 metres tall. And we’re also told that the observer is standing 12 metres away. So, we have these two sides of the right-angled triangle. The angle of elevation, remember, is measured from the horizontal to the top of the palm tree. And therefore, it is this angle here that we’re looking to calculate.

So, for any problem involving trigonometry, it’s always a good idea once you’ve got the diagram to then label the three sides of the triangle. So, in relation to this angle 𝜃, I’m going to give them their labels of the opposite, the adjacent, and the hypotenuse. And once I’ve done that, I can see that the two lengths I have are the lengths of the opposite and the adjacent sides. Now, this tells me that it’s the tan ratio that I’m going to be using in order to calculate this angle. Because if you think back to SOHCAHTOA, then TOA is where the opposite and the adjacent appear together.

So, I recall the definition of the tan ratio. And then, I need to write it down using the information in this question. So, I’m going to replace the opposite with 10.6 and the adjacent with 12. I have then that tan of 𝜃 is equal to 10.6 over 12. Now, I’m looking to calculate that angle 𝜃, so I need to use the inverse tan function here. So, I have then that 𝜃 is equal to tan inverse of this ratio, 10.6 over 12. Now, I can use my calculator in order to evaluate that, making sure my calculator is in degree mode. And this tells me that 𝜃 is equal to 41.4552.

Now, this answer is in degrees. And I’ve been asked in the question to give my answer to the nearest minute. So, I now need to convert this from degrees into degrees and minutes. So, what I can see then is that I have 41 full degrees. And then, I also have this decimal 0.4552 and so on left over. And that’s the part that I need to convert into minutes. Now, remember that a minute is one sixtieth of a degree. So, in order to work out what this decimal represents in terms of minutes, I need to multiply it by 60.

When I do so, this gives me an answer of 27.3140. So, rounding that to the nearest minute, I have 27 minutes. Finally, I need to pull these two parts of my answer together. And doing so tells me that this angle of elevation is 41 degrees 27 minutes to the nearest minute.

So, in this question then, we drew a diagram using the information that we were given. We identified the need for the tangent ratio because we had the opposite and adjacent sides. We then used the inverse tangent function in order to work out the size of this angle, finally converting our answer into degrees and minutes.

This problem is about a circle. It tells us that a chord is drawn in a circle whose radius is eight centimetres. If the size of its central angle is 100 degrees, we’re asked to calculate the length of the chord to the nearest thousandth.

So, as I haven’t been given a diagram, I’m gonna start off by drawing my own. And I’m gonna begin with a circle. Now, it tells me that a chord is drawn in the circle, so that is a line connecting two points on the circumference but not passing through the center of the circle. So, here is my chord. And then, it tells me that the central angle is a 100 degrees. So, I’m gonna connect the two ends of the chord to the center of the circle. And that’s going to create an angle of 100 degrees.

Now, there’s some other information we know, which is that the radius of the circle is eight centimetres. So, those two lines connecting the center of the circle to the ends of the chord are both eight centimetres. So, what I can see then is I have an isosceles triangle because this triangle has two sides the same length. In order to do trigonometry though, I’d like to have a right-angled triangle.

So, what I’m going to do is I’m gonna draw in the perpendicular height of this isosceles triangle. And what that does is it divides the triangle up into two congruent, that is identical, right-angled triangles. So, that’s drawing in this line here. Now, because that line divides the isosceles triangle directly in half, that 100-degree angle is now split evenly, 50 degrees and 50 degrees, into each of these two right-angled triangles. So, to help you visualize it a little bit better, I’m just gonna draw out one of these two right-angled triangles. So, here we have then that triangle a little bit bigger.

Now, I’m asked to calculate the length of the chord. So, what I’m going to do is I’m going to use trigonometry to calculate the base of the triangle, this side here. And then I’m going to double it in order to get the full length of the chord. So, the first step then is to label all of the sides of the triangle in relation to this angle of 50 degrees. So, we have the opposite, the adjacent, and the hypotenuse. So, what I can see then is that I’m going to be using the sine ratio because I know the hypotenuse and I want to calculate the opposite, and so that’s the SOH part of SOHCAHTOA.

Now, I’ll just give the opposite a different letter so that we can save confusion with zero in our working out. I’ll call it 𝑥 centimetres. So, we recognized the need for the sine ratio. And I’ve recalled its definition here. It’s that sin of 𝜃 is equal to the opposite divided by the hypotenuse. So, I’m gonna write this ratio out using the information in this question. I’m gonna replace the angle with 50 degrees, the opposite with 𝑥, and the hypotenuse with eight. So, then, here is an equation that I can use in order to work out the value of 𝑥.

Now, 𝑥 is currently divided by eight. So, I’m gonna multiply both sides of this equation by eight. Now, I’ve just swapped the two sides of the equation round here. But this tells me that 𝑥 is equal to eight sin 50. Now, I could evaluate it at this stage here. But actually, I’m not asked to find 𝑥, remember, I’m asked for the length of the chord. So, I need to double this. So, what I’m gonna do is just multiply this by two. So, I have that the length of the chord is two lots of eight sin 50, which is just equivalent to 16 sin 50.

And now, I’m going to use my calculator to evaluate this. And it tells me that this is equal to 12.2567. Now, the question asked me to give my answer to the nearest thousandth. So finally, I’ll round my answer. And therefore, I have that the length of this chord to the nearest thousandth is 12.257 centimetres.

In this question, we’re given a triangle and we can see that it’s an isosceles triangle because we’re told that 𝐴𝐵 is equal to 𝐴𝐶. We’re also given some other information and we’re asked to find the area of this triangle.

First of all, then, I’m told about the value of cos of 𝐵. But there are currently no right-angled triangles visible within the question. So, what I want to do is create a right-angled triangle. What I can do, because triangle 𝐴𝐵𝐶 is isosceles, if I draw in the perpendicular height of this triangle, then it would divide this isosceles triangle into two congruent right-angled triangles. So, that’s filling in this height here. Now, that also means that 20 centimetres will be split exactly into two lots of 10 centimetres.

Now, I know I’m going to need trigonometry in this problem. So, within the right-angled triangle on the left, I’m just gonna label the three sides in relation to that angle 𝐵. So, we have the opposite, the adjacent, and the hypotenuse. Now, I’m told that cos of 𝐵 is equal to five over 13. So, I’m also going to need the definition of the cosine ratio. And remember, it’s that the cos of a general angle 𝜃 is equal to the adjacent over the hypotenuse.

So, let’s think about a strategy for this question. I need the area of triangle 𝐴𝐵𝐶. Well, I know the base is 20 centimetres. So, I also need to know the perpendicular height cause then I can work out the area by doing base multiplied by perpendicular height over two. So, this means I need to work out the length of the opposite side. I haven’t been given enough information in order to work out the opposite side straight away because the information about the cosine ratio is about the adjacent and the hypotenuse.

My strategy then, it’s going to be to work out the length of the hypotenuse using trigonometry and then use the Pythagorean theorem, which we’ll look at in little bit more detail later, in order to find the value of this third side.

So, let’s start off by writing down the cosine ratio for this triangle then. We have that the cos of 𝐵 is equal to 10 over 𝐻. And we also know that this is equal to five over 13. So, I have here an equation that I can solve in order to work out the value of 𝐻. Now, 𝐻 is in the denominator. So, I’m gonna multiply both sides of the equation by 𝐻. And I’m also gonna multiply both sides of the equation by 13 at the same time. Now, when I do that, I get that 130 is equal to five 𝐻. Next, I want to divide both sides of this equation by five. And doing so tells me that 𝐻 is equal to 26. So, I’ve got the length of the hypotenuse of this triangle.

So, I said we’re going to use the Pythagorean theorem to work out the length of the third side. So, let’s just recall that. So, it’s often referred to as 𝑎 squared plus 𝑏 squared is equal to 𝑐 squared. But what it tells us is if I take the two shorter sides of a right-angled triangle, square them, and add them together, then I get the same result as if I square the longest side, the hypotenuse.

So, I’m gonna apply the Pythagorean theorem here. And I’m gonna refer to the opposite side as 𝑥 rather than 𝘖 just to avoid mixing up with zero. I have then that 𝑥 squared plus 10 squared is equal to 26 squared. Now, I’ve written in the complete working out here. And you can pause the video to look at it in detail if you want. But essentially, I’m evaluating 10 squared and 26 squared, subtracting a 100, square rooting, and this tells me that 𝑥 is equal to 24.

So finally, then, I have all the information I need in order to find the area of this triangle. I have the full base of the isosceles triangle, 20 centimetres, and the perpendicular height, 24 centimetres. So, the area is 20 multiplied by 24 and then divided by two. We have then a final answer of 240 centimetres squared.

In this question, we have a diagram that involves two right-angled triangles. And we’re asked to calculate the length of 𝐴𝐶, which is a side in the larger right-angled triangle.

So, first of all, let’s just consider what we’ve got. In the smaller right-angled triangle, triangle 𝐶𝐷𝐸, I’ve been given two lengths. And in the larger right-angled triangle, triangle 𝐴𝐵𝐶, I have only one length. And it’s the length in that triangle that ultimately I’m looking to calculate. Now, there is an angle that is common to both of these triangles. And it’s angle 𝐶. So, what my approach is going to be is to calculate angle 𝐶 first using the smaller right-angled triangle and then combine angle 𝐶 with that length 19 centimetres in the larger triangle in order to work out the length of 𝐴𝐶.

So, starting off in that smaller right-angled triangle first, I’m gonna label the three sides in relation to that angle 𝐶. So, we have the opposite, the adjacent, and the hypotenuse. Now, what I can see is that because I have the length of the opposite and the hypotenuse, I can use the sine ratio in order to calculate this angle 𝐶. The sine ratio, remember, is defined as the opposite divided by the hypotenuse. So, this tells me that sin of 𝐶 is equal to 11 over 21 in this case.

Now, I could evaluate angle 𝐶 at this point by using the inverse sine function. And if I did, it would tell me that angle 𝐶 is equal to 31.6 degrees. But actually, that’s not necessary. And you’ll see why a little bit later in the question. So, now, let’s consider the large triangle, triangle 𝐴𝐵𝐶. And, again, I’m gonna label its three sides in relation to angle 𝐶. So, we can see that we have the length of the opposite side, 19 centimetres. And the length we want to calculate, 𝐴𝐶, is the hypotenuse. So, again, we’re going to be using the sine ratio.

So, if I now write down the sine ratio in this larger triangle, well, we have then that sin of this angle 𝐶 is equal to 19 over 𝐴𝐶, that side. So, what I can do then is I can put these two pieces of information together. As they’re both equal to sin 𝐶, I can then equate them to each other. And this tells me then that 11 over 21 is equal to 19 over 𝐴𝐶. That’s why I didn’t actually need to evaluate the angle 𝐶 because it was sin 𝐶 that I use later in the question.

Now, I have an equation that I can solve in order to work out the value of 𝐴𝐶. So, I’m gonna multiply both sides of the equation by 𝐴𝐶 and also by 21. So, this tells me that 11 lots of 𝐴𝐶 is equal to 19 multiplied by 21. Next, I’m gonna divide both sides of the equation by 11. So, I have that 𝐴𝐶 is equal to 19 multiplied by 21 over 11. And I’ll give my answer for 𝐴𝐶 as a fraction. We have then that 𝐴𝐶 is equal to 399 over 11. And the units for that are centimetres.

In summary then, we’ve recalled the definition of the three trigonometric ratios sine, cosine, and tangent. We’ve seen how to identify which ratio is required. And then, we’ve applied them to some mixed problems.