# Lesson Video: Right Triangle Trigonometry: Solving for an Angle

Learn the definitions of the inverse sine, cosine and tangent ratios and how to apply them to calculate angles in right-angled triangles when two side lengths are given.

12:11

### Video Transcript

In this video, we’re going to see how to apply the inverse of the three trigonometric ratios, sine, cosine, and tangent, in order to calculate angles in right-angled triangles.

First of all, let’s define these inverse trigonometric ratios. I have here a diagram of a right-angled triangle in which I’ve labeled one of the angles as 𝜃. I’ve then labeled the three sides of the triangle in relation to the angle 𝜃, so we have the opposite, the adjacent, and the hypotenuse. The three trigonometric ratios, sine, cosine, and tangent, are the ratios that exist between different pairs of sides in this right-angled triangle.

So, the sine ratio, sin of angle 𝜃, is the opposite divided by the hypotenuse. Cosine ratio is cos of 𝜃 is the adjacent divided by the hypotenuse. And finally, tangent, the tan ratio, is the opposite divided by the adjacent. A useful way to remember these is to recall the word SOHCAHTOA, where each of those letters represents the first letter in each of these words. So, CAH, for example, the C is for cos, the A is for adjacent, and the H is for hypotenuse. So, CAH tells us that the cos ratio is adjacent divided by hypotenuse.

Those ratios are the trigonometric ratios, and they’re particularly useful if we know a side and an angle and are looking to calculate another side. But in this video, we’re looking at how to calculate angles. And therefore, we need what’s referred to as the inverse trigonometric ratios. These are defined like this. The notation we use is sin, cos, or tan and then a superscript negative one, which is said as sine inverse or inverse sine. And what they mean are, if I know the value of the ratio of opposite divided by hypotenuse, then I can work backwards using this inverse sine function in order to calculate the angle that that ratio is associated with.

So, when we know two sides of a right-angled triangle, we can use the relevant inverse trigonometric ratio in order to calculate an angle. On your calculator, you will see that above the sin, cos, and tan button, there is usually sin inverse, cos inverse, and tan inverse. You often have to press shift in order to get to these functions. We’ll now look at a couple of examples of how to apply these inverse trigonometric ratios.

For the given figure, find the measure of angle 𝜃 in degrees to two decimal places.

So, we have a diagram of a right-angled triangle, and we can see that we’re given the lengths of two of the sides of this triangle. They’re three units and eight units. And we’re looking to find the size of this angle 𝜃.

As we’re using trigonometry for this problem, the first step is going to be to label all three of the sides of the triangle in relation to their angle 𝜃. So, we have the opposite, the adjacent, and the hypotenuse. We can see then that the two sides of the triangle we have are the adjacent and the hypotenuse. If I think back to the acronym of SOHCAHTOA, then A and H appear together in the CAH part, which tells me that it’s the cosine ratio I’m going to need to use in this question.

The definition of the cosine ratio, remember, was that cos of the angle 𝜃 is equal to the adjacent divided by the hypotenuse. So, I’m gonna write down this ratio using the information in this particular question. And therefore, I have that cos of 𝜃 is equal to three over eight. Now, this is where I need to use the inverse trigonometric function. If cos of 𝜃 is equal to three over eight, then 𝜃 is equal to cos inverse of three over eight.

At this stage, I use my calculator to evaluate this, remembering that that cos inverse button is usually directly above the cos button. This tells me that 𝜃 is equal to 67.97568 degrees. The question asked me to round my answer to two decimal places. Therefore, my final answer is that 𝜃 is equal to 67.98 degrees. So, in this question, we identified the need for the cosine ratio because the lengths we were given were the adjacent and the hypotenuse. We wrote down that ratio using these lengths. We then used the inverse cosine ratio in order to calculate the value of this angle 𝜃.

Find the measure of angle ACB giving the answer to the nearest second.

I’ve got a diagram of a right-angled triangle, and I’m asked to find the measure of angle ACB. So, that means the angle formed when I move from A to C to B. It’s this angle here that I’m looking for, so I’ve given it the label 𝜃. As we’re going to use trigonometry to solve this problem, I’m gonna label the three sides of the triangle with their names in relation to that angle of 𝜃. So, we have the opposite, the adjacent, and the hypotenuse.

Now, I can see that the two sides I’ve been given are the opposite and the adjacent. If I think back to that acronym SOHCAHTOA, then I see that it’s the tan ratio I’m going to need because the opposite and the adjacent appear together in the TOA part. So, the definition of the tan ratio is the opposite divided by the adjacent. For this triangle, then, that is 43 divided by 26. So, we have tan of 𝜃 is equal to 43 over 26. Now, I need to use the inverse tangent ratio.

So, if tan 𝜃 is 43 over 26, then 𝜃 is equal to tan inverse of 43 over 26. I use my calculator to evaluate this. And it tells me that 𝜃 is equal to 58.84069. Now, this is an answer in degrees, and the question has asked me to give my answer to the nearest second. So, I need to recall how to convert a value from degrees into degrees, minutes, and seconds. So, I have 58 full degrees, and then I have this decimal of 0.840695 and so on, which needs to be converted into minutes and then seconds.

Remember that a minute is one sixtieth of a degree. So, to work out what this decimal represents in minutes, I need to multiply it by 60. When I do that, I get this value of 50.4417295. This tells me that there are 50 full minutes and a decimal of 0.4417295, which needs to be converted into seconds. A second is one sixtieth of a minute. So, again, to convert this into seconds, I need to multiply it by 60. When I do this, I get a value of 26.5037. So, in order to round this to the nearest second, I round it up to 27 seconds.

Finally, I need to pull the three parts of this answer together. And in doing so, my final answer then is that the measure of angle ACB is 58 degrees, 50 minutes, 27 seconds to the nearest second. So, within this question, we identified the need for the tan ratio, because the lengths we were given were the opposite and the adjacent. We wrote down the tan ratio for this triangle. We then applied the inverse tan ratio to work out the angle that this ratio was associated with and finally converted this answer from degrees into degrees, minutes, and seconds.

A car is going down a ramp which is 10 metres high and 71 metres long. Find the angle between the ramp and the horizontal, giving the answer to the nearest second.

So, this question is a worded problem, and we haven’t been given a diagram. I would always suggest that, in this situation, you draw your own diagram to start off with. So, here we have an idea of what this ramp might look like. It is 71 metres long and 10 metres high. We’re asked to find the angle between the ramp and the horizontal, so we’re looking to calculate this angle here, which I’ve labeled as 𝜃.

Now, we’re going to solve this problem using trigonometry. So, I’m going to begin by labeling the three sides of this triangle in relation to this angle 𝜃. Having done this, I can see that the two lengths that I’ve been given represent the opposite and the hypotenuse of this right-angled triangle. Thinking back to that acronym of SOHCAHTOA, this tells me that it’s the sine ratio I’m going to need in this problem, as O and H appear together in the SOH part of SOHCAHTOA.

So, the definition of the sine ratio is that it’s the opposite divided by the hypotenuse. I then write this ratio down for this particular triangle, and I have then that sin of 𝜃 is equal to 10 over 71. In order to work out the size of this angle, I need to apply the inverse sine ratio. So, if sin of 𝜃 is 10 over 71, then 𝜃 is sin inverse of 10 over 71. I evaluate this with my calculator, and I see that 𝜃 is equal to 8.09674977.

Now, this is an answer in degrees, and the question has asked me to give my answer to the nearest second. So, let’s recall how to convert an answer from degrees into degrees, minutes, and seconds. I have eight full degrees, and then I have a decimal left over of 0.09674977, which needs to be converted first into minutes and then into seconds. Now, recall that a minute is one sixtieth of a degree. So, in order to work out what this decimal represents in minutes, I need to multiply it by 60. This gives me a value of 5.804986188.

What this tells me then is that I have five full minutes and then a decimal of 0.80 and so on left over. This decimal needs to be converted into seconds. So, recall that a second is one sixtieth of a minute. And therefore, in order to see what this decimal represents in seconds, I need to multiply it by 60. When I do this, I get a value of 48.29917126. And if I round that to the nearest second, it’s 48 seconds. Finally, then, I need to combine the three parts of my answer. And this tells me that the angle between the ramp and the horizontal, to the nearest second, is eight degrees, five minutes, and 48 seconds.

So, within this question, we drew our own diagram as we weren’t given one within the question itself. We identified the need for the sine ratio because the two lengths we have been given were the lengths of the opposite and the hypotenuse. We used the inverse sine ratio to calculate the angle that was associated with that value. And then, we converted our answer from degrees into degrees, minutes, and seconds.

In summary, then, the three inverse trigonometric ratios sine inverse, cosine inverse, and tan inverse can be used to calculate an angle in a right-angled triangle when we know at least two of the sides. The ratio that we choose will depend on which two sides we’re given in exactly the same way as it does when we’re calculating the length.