Trigonometric Functions on Right Triangles: Videos & Practice Problems

Trigonometric functions, including sine, cosine, and tangent, relate angles to side lengths in right triangles. The sine of an angle is the ratio of the opposite side to the hypotenuse, while cosine is adjacent over hypotenuse, and tangent is opposite over adjacent. Special triangles, like the 45-45-90 and 30-60-90 triangles, have specific ratios that simplify calculations. For example, in a 45-45-90 triangle, both legs are equal, and the hypotenuse is the leg length multiplied by the square root of 2. Understanding these relationships aids in solving trigonometric problems efficiently.

Introduction to Trigonometric Functions

Hey, everyone, and welcome back. So in earlier videos, we've brushed up on this idea of angles and triangles. And what we're going to be looking at in this video is arguably one of the most critical topics, both in this course and in future math courses, that you will likely take, which is the idea of trigonometric functions, which we also call trig functions for short. Now, if you're not sure what trig functions are, all these do is they relate angles to the side lengths of a right triangle. Now in this video, there are three main types of trig functions that we're going to talk about, which are the sine, cosine, and tangent.

Now that might sound a little confusing or like it's a bunch of gibberish, but don't sweat it, because all we're going to learn is that these three trigonometric functions are just ratios. And an example of a ratio would be something like three fifths or four fifths. So these types of fractions or ratios are all that these trigonometric functions are. Without further ado, let's get into this video and see how we can solve some problems where we need to find these specific ratios based on a right triangle. Now, how can we go about finding the sine of some angle in this triangle that we'll call theta?

How can we find the sine of this? Well, sine is opposite over hypotenuse. So if you wanted to find the sine of this angle theta, what you do is look at the angle that you have and go to the opposite side of the triangle, which in this case is three, and then divide that by the hypotenuse, which is always the long side of the triangle or five. So three fifths is going to be the sine of theta. As you can see, solving for these trigonometric functions is pretty straightforward.

So let's now say we wanted to find the cosine of theta. Well, if we're dealing with the cosine this time, what we need to do is look at our angle theta and find the adjacent side and divide it by the hypotenuse. Now the adjacent side of this triangle is the side next to the angle, which in this case is four, and then this is divided by the hypotenuse - the long side of the triangle, which is five. So the cosine of our angle theta is four fifths. But now let's say we wanted to find the tangent of our angle theta.

If we're dealing with the tangent this time, what we need to do is take the opposite side of the triangle and divide it by the adjacent side. So the opposite side to our angle theta is three, and then the adjacent side is four. So 34 would be the tangent of our angle theta. Now, you might feel like this is a lot to remember. Like, how are we just supposed to know that sine is opposite over hypotenuse, or that tangent is opposite over adjacent?

Well, it turns out there is a memory tool that we use to remember this, and that is SOHCAHTOA. Because this memory tool, SOHCAHTOA, tells us how each of the trigonometric functions relates to the sides of the right triangle.
So we have that sine is opposite over hypotenuse,
cosine is adjacent over hypotenuse,
and tangent is opposite over adjacent.
Now one more thing I want to mention about the tangent is that the tangent of theta is also equal to the sine of theta divided by the cosine of theta.
So if we were to take the sine of theta, which we figured out was three fifths, and divide that by the cosine of theta, which we figured out was four fifths, notice how the fives would just cancel in this fraction.

So all we would end up with is 34. And 34 is exactly what we calculated the tangent to be. So this is another relationship to be familiar with when dealing with these trigonometric functions. Now to really make sure we have these trig functions and this SOHCAHTOA memory tool down, let's try some examples where we have to solve some problems based on a right triangle. Now in this problem, we are asked to find the values of the trig function indicated given the right triangle that we have over here.

Now we're going to start with example a. And before I do anything, what I'm going to do is write down the memory tool SOHCAHTOA. Now looking at example a, we have the sine. And sine is opposite over hypotenuse according to SOHCAHTOA. So we'll have the opposite side of our right triangle divided by the hypotenuse.

Now we're dealing with angle x in this example, so if we go to angle x in our triangle, we're going to have the side opposite to angle x, which is 12, divided by the hypotenuse or the long side, which is 13. So 12 over 13 is the ratio we're looking for and the answer for example a. Now let's try solving example b, where we are asked to find the tangent of x. Well, if we once again go to our angle x here, we can find the tangent by looking at SOHCAHTOA, which says that tangent is opposite over adjacent. So if we take this ratio, we can go to our angle x, and we can find the side opposite to x, which in this case is 12, and then divide it by the adjacent side or the side next to x, which in this case is five.

So 12 over five is the ratio that we're looking for, for example b. But now let's try solving example c, where we are asked to find the cosine of our angle y. Now when dealing with the cosine, cosine is adjacent over hypotenuse based on this memory tool SOHCAHTOA. Now what you might think we need to do is look at our angle and then find the adjacent side, which would be five, and divide it by the hypotenuse or the long side, which is 13. Now if five thirteenths is what you're thinking the correct ratio is, this is actually not correct, and here's the reason why.

Notice in this example, we are dealing with angle y. And because we're dealing with angle y, we can no longer use x as a reference. We now have to use y as the reference angle. So rather than having five over 13, we're going to have the ratios based on the adjacent and hypotenuse of y. So if we go to y, the adjacent side would actually be 12, and then the hypotenuse or the long side is still 13.

So 12 over 13 would be the ratio for the cosine of y. So this is something you need to watch out for when dealing with these types of problems. Because notice in the first two examples we were dealing with angle x, so we used x as a reference. Where in this example, we were dealing with y, so we used angle y as a reference. So that's the main idea behind solving trigonometric functions and using this SOHCAHTOA memory tool.

Hope you found this video helpful. Thanks for watching, and please let me know if you have any questions.

Given the right triangle below, evaluate $\cos\theta$.

• A. $\cos\theta=\frac{8}{17}$
• B. $\cos\theta=\frac{8}{15}$
• C. $\cos\theta=\frac{15}{17}$
• D. $\cos\theta=\frac{15}{8}$

Answer: (C)

Given the right triangle below, evaluate $\tan\theta$.

• A. $\tan\theta=\frac35$
• B. $\tan\theta=\frac45$
• C. $\tan\theta=\frac43$
• D. $\tan\theta=\frac34$

Answer: (D)

Fundamental Trigonometric Identities

Hey everyone. So in the last video, we learned about the three main types of trigonometric functions, which are the sine, cosine, and tangent. And what we're going to be learning about in this video is three new trigonometric functions known as the cosecant, secant, and cotangent. Just like the three trig functions we learned about in the previous video, these functions also have abbreviated versions of their names. We have the cosecant, secant, and cotangent.

This might sound really scary and kind of complicated because there's just a lot to remember and a lot of different things that we've learned about so far, but don't sweat it with these three trigonometric functions. Because what we're going to learn in this video is that these three functions are actually very closely related to the last three trigonometric functions we learned about in a very simple way. So without further ado, let's get right into things to see what these new trig functions are all about. These functions are reciprocals of the other trig functions. Specifically, we call these reciprocal identities.

A reciprocal identity is just a fancy way of saying that this thing is a flipped version of that thing. To understand this a little bit better, let's take a look at the cosecant for example. The cosecant is a reciprocal identity for the sine. That means the cosecant is just a flipped version, meaning cosecant would be hypotenuseopposite. So, if the sine of theta is oppositehypotenuse, the cosecant of theta would be hypotenuseopposite.

Going to a right triangle, for example, we can look at our angle theta, and the hypotenuse or the long side would go on top of the fraction, and then the side opposite to the angle, which in this case is three, would go on the bottom of the fraction. That's the idea of the cosecant. The secant is a reciprocal identity for the cosine. That means that the secant of theta is 1cosine. So if cosine is adjacenthypotenuse, then secant is hypotenuseadjacent.

Going to our triangle, we have the hypotenuse or long side, which is five, and this is divided by the side adjacent to our angle theta, which in this case is four. That's the secant of theta. It turns out that the tangent of theta also has a reciprocal identity, which is the cotangent of theta. Cotangent of theta would be 1tangent. So if the tangent of theta is oppositeadjacent, then the cotangent of theta would be adjacentopposite.

Notice how the fractions for the cosecant, secant, and cotangent are literally just the flipped versions of the sine, cosine, and tangent. This is the main idea and concept behind reciprocal identities. Another identity to watch out for when dealing with these types of trigonometric functions is how the cotangent of theta is cosine over sine. So this same rule of reciprocal identities applies here as well.

To really make sure we have this concept solidified, let's actually try some examples where we have to use these new reciprocal identities. In these problems, we're told to find the value of the trig function indicated given this right triangle. Let’s start by taking a look at example a. For example a, we are asked to find the secant of our angle x. When dealing with the secant, we learn the secant is related to the cosine. The cosine of x is going to be equal to adjacenthypotenuse.

If we go to our triangle and look at our angle x, the side adjacent to x is five and this is divided by the hypotenuse or the longest side which is 13. So this is what the cosine is equal to and since the secant is just the reciprocal of the cosine, we just flip this fraction. It would become 135. That is how you can find the secant of angle x using these reciprocal identities. But now let's take a look at example b, where we are asked to find the cosecant of x. The cosecant is related to the sine. The sine of x is going to be equal to oppositehypotenuse. We'll go into our angle x, we can see the side opposite to x is 12, and then this is divided by the hypotenuse or the long side, which is 13. If we wanted to find the cosecant of x, which is just the flipped version of sine, we just flip this fraction and get 1312. That is the cosecant of our angle x.

Now for example c, we're asked to find the cotangent of y, but notice this time we're looking at angle y, so we can no longer be with reference to x. You want to make sure you are paying attention to which angle you are with reference to. Since we were with reference to angle y, we can find the cotangent of y by recognizing that the cotangent is the reciprocal for the tangent. The tangent of y we can see here is equal to oppositeadjacent.

We'll go into angle y here the side opposite to angle y is five, then this is divided by the side adjacent to y which is 12. The flipped version of five over 12 would be 125 which is equal to the cotangent of y. So this is the main idea behind these three new trigonometric functions and how you can solve problems with them using these reciprocal identities. I hope you found this video helpful.

Thanks for watching, and please let me know if you have any questions.

If $\tan\theta=\frac{12}{5}$, find the values of the five other trigonometric functions. Rationalize the denominators if necessary.

• A. $\sin\theta=\frac{12}{13},\cos\theta=\frac{5}{13},\cot\theta=\frac{5}{12},\sec\theta=\frac{13}{5},\csc\theta=\frac{13}{12}$
• B. $\sin\theta=\frac{5}{13},\cos\theta=\frac{12}{13},\cot\theta=\frac{5}{12},\sec\theta=\frac{13}{12},\csc\theta=\frac{13}{5}$
• C. $\sin\theta=\frac{12}{13},\cos\theta=\frac{5}{13},\cot\theta=-\frac{5}{12},\sec\theta=-\frac{13}{5},\csc\theta=-\frac{13}{12}$
• D. $\sin\theta=\frac{5}{13},\cos\theta=\frac{12}{13},\cot\theta=-\frac{5}{12},\sec\theta=-\frac{13}{12},\csc\theta=-\frac{13}{5}$

Answer: (A)

If $\sin\theta=\frac{\sqrt{17}}{17}$, find the values of the five other trigonometric functions. Rationalize the denominators if necessary.

• A. $\cos\theta=\frac{\sqrt{17}}{4},\tan\theta=\frac14,\cot\theta=4,\sec\theta=\sqrt{17},\csc\theta=\frac{4\sqrt{17}}{17}$
• B. $\cos\theta=\frac{\sqrt{17}}{4},\tan\theta=-\frac14,\cot\theta=-4,\sec\theta=\sqrt{17},\csc\theta=\frac{4\sqrt{17}}{17}$
• C. $\cos\theta=\frac{4\sqrt{17}}{17},\tan\theta=-\frac14,\cot\theta=-4,\sec\theta=\frac{\sqrt{17}}{4},\csc\theta=\sqrt{17}$
• D. $\cos\theta=\frac{4\sqrt{17}}{17},\tan\theta=\frac14,\cot\theta=4,\sec\theta=\frac{\sqrt{17}}{4},\csc\theta=\sqrt{17}$

Answer: (D)

How to Use a Calculator for Trig Functions

Hey, everyone. So up to this point, we've been spending a lot of time talking about trigonometric functions as well as inverse trigonometric functions. And in this video, we're going to be tying these concepts together by talking about how we can use a calculator to solve problems that involve these types of functions. Now, something that I will say is that for certain problems, you're actually going to have to use your calculator to evaluate the function rather than just using the fraction or ratios that we learned about before. This can be pretty complicated, especially when you're not too sure how to use your calculator properly, but that's what we're going to be going over in this video.

And I'm going to be walking you through the steps that you need to take to evaluate these types of trigonometric functions and solve these problems. So, without further ado, let's get right into things. Now for trig functions, you're going to need to use the sine, cosine, and tangent buttons on your calculator, which should look something like this. Now when dealing with problems where you have to evaluate the trig function on a calculator, you need to make sure you're in the correct mode. The mode button is something that should also be on your calculator.

And when you press the mode button, you'll be taken to a certain menu. And on this menu, you will have an option to choose between degrees or radians. Now, depending on which of these you choose is dependent on the type of problem that you're solving, but you need to know when to be in the correct mode for the type of problem. So, let's actually take a look at some examples to make sure that we know how to do this. So here we have an example which says find the value of each of the following trigonometric operations and round to the nearest tenth.

And keep in mind that the nearest tenth would be the nearest number or the first number after the decimal place. Now let's start with example a, where we're asked to find the sine of 37 degrees. Now notice in this example, we have a degree sign. And because we have degrees, that means that we want to be in degree mode on our calculator. So once you switch to degree mode, you're going to type in the sine, which is a button on your calculator, and then what you're going to do is put 37 within these parentheses.

You're going to close the parentheses and then hit enter. Once you click enter on your calculator, rounded to the nearest tenth, you should get about 0.6. So 0.6 would be the decimal approximation for what you get from the sine of 37 degrees, and that's example a. But now let's take a look at example b, where we're asked to find the tangent of two pi over 15. Now notice in this example, we're dealing with pi.

And because of this, that means you actually want to be in radian mode for this example. So go to mode on your calculator and switch to radians. Once you've done this, you're going to find the tangent button on your calculator, and then you're going to type in two pi, and then divide that by 15. You're then going to close the parenthesis and press enter. Once you press enter, you should get about 0.4 rounded to the nearest tenth.

So that's going to be the answer for example b. Now let's take a look at example c, where we're asked to find the secant of 50 degrees. I again notice that we have a degree symbol here, which means we want to be in degree mode for this example. Now the problem with this situation is we don't really have a secant button on our calculator. But if you recall from the reciprocal identities, secant is the same thing as one over cosine.

So this would be the same thing as taking one and dividing it by the cosine of 50 degrees. So what I'm going to do is put in one into my calculator and then divide it by the cosine of 50. You can then close that parenthesis, and then you can hit enter. And when you hit enter on the calculator, you should get an approximate value of 1.55, but rounded to the nearest tenth, we can say that that's 1.6. So 1.6 would be the answer for example c.

Now let's try example d, where we're asked to find the arctangent of three fourths. We are asked to find our answer in degrees, so we want to be in degree mode on our calculator. But the question becomes, how can we find the arctangent of something? Well, we actually discussed this in previous videos. The arctangent is the same thing as the inverse tangent.

And whenever you're dealing with an inverse trig function, what you want to do is you want to press the second button on your calculator. This is a button that you should see. Once you press the second button, you're then going to press the associated trigonometric function, and this should give you the inverse of that trig function. So what you want to do is go to your calculator, and you want to press the second button. And then once you've pressed second, you're going to press on tangent.

This will give you the inverse tangent. And from here, you can type in three, divide it by four, and then close the parenthesis. Once you've done this, you can hit Enter on this, and you should get an approximate value of 36.9 degrees rounded to the nearest tenth. So 36.9 degrees is the answer for example d. So this is how you can evaluate trigonometric and inverse trigonometric functions on your calculator.

So, hope you found this video helpful. Thanks for watching, and let me know if you have any questions.

What is a positive value of A in the interval $\left[0\degree,90\degree\right)$ that will make the following statement true? Express the answer in four decimal places.
$\sin A=0.9235$

• A. $22.5568°$
• B. $67.4432°$
• C. $22.4432°$
• D. $33.5438°$

Answer: (B)

What is the positive value of P in the interval $\left[0\degree,90\degree\right)$ that will make the following statement true? Express the answer in four decimal places.
$\cot P=5.2371$

• A. $55.8102°$
• B. $34.1898°$
• C. $10.8102°$
• D. $79.1898°$

Answer: (C)

What is the positive value of $D$ in the interval $\left[0,\frac{\pi}{2}\right)$ that will make the following statement true? Express the answer in four decimal places.
$\sec D=3.2842$

• A. 0.3094 rad
• B. 1.2614 rad
• C. 0.4760 rad
• D. 1.0934 rad

Answer: (B)

How to Use a Calculator for Trig Functions Example 1

Welcome back, everyone. Let's see if we can solve this example. In this example, we are asked to determine the missing angle theta in degrees for the right triangle below and to approximate our answer to two decimal places. Now, the example we have here is a continuation of an example we did in a previous video where we learned about inverse trigonometric functions. Notice that we have these first three steps complete because we've already chosen a trig function that includes the correct sides and angles, which is the sine.

We've gone ahead and written the equation, sinθ=ah, and we've taken the inverse on both sides of the equation to solve for our angle theta. So what we need to do is now approximate what this angle theta is based on what we've solved so far. So let's go ahead and see how we can do this. Now, what I notice is, in this example, we are asked to give our answer in degrees. So because of this, you want to make sure your calculator is in degree mode when solving this problem.

Once you've put your calculator in degree mode, the next step is going to be to press the second key on your calculator and the associated trig function. This will give you the inverse trig function you're looking for. And since I can see we're looking for inverse sine, you're going to want to press the second button and then sine on your calculator. Now once you have the inverse sine, you're going to type in the ratio 613 and then close that parenthesis. From here, you're going to click enter, and this will give you the approximate value for the inverse trig function, and we're asked to round our answer to two decimal places.

Now, looking at the calculator, your angle theta should be approximately equal to 27.49 degrees. So this right here is the approximate answer for our angle theta and the solution to this problem. Hope you found this video helpful. Thanks for watching.

45-45-90 Triangles

Welcome back, everyone. So up to this point, we've spent a lot of time talking about trigonometric functions, the Pythagorean theorem, and how they all relate to the right triangle. Now what we're going to be learning about in this video is some of the special and common right triangles that you're going to see. And in this video specifically, we're going to be talking about the 45-45-90 special triangle. Now the reason that these triangles are special is because it turns out that because they show up relatively frequently, there are actually some shortcuts that you can use to solve these triangles very fast.

So if you don't like all the brute force work we've been doing with trigonometric functions and the Pythagorean theorem, you're going to learn some shortcuts for solving these triangles in this video. So without further ado, let's get right into things. Now when you have a triangle with 45-degree angles, like this triangle down here, for example, this is going to be a situation where you have the special 45-45-90 triangle. And in these triangles, the two legs of the triangle are always going to be the same length. So if you ever see a situation where you have a right triangle and two of the legs are the same, that means you're dealing with this special triangle.

Now what we can do with this is we can actually solve for the hypotenuse of the triangle by simply taking a multiple of the leg length. And the multiple that you're going to look for is the square root of two. Because if you take a leg, like five, and you multiply it by the square root of two, this will give you the hypotenuse. And that's the answer. We just solved for the long side of this triangle.

So as you can see, this shortcut right here makes solving for the sides of the triangle really straightforward and really fast. Now if you didn't remember this relationship, there is another strategy you can use, which is simply the long version of using the Pythagorean theorem. So let's say that we set this side to a, that side to b, and then the hypotenuse equal to c, and we want to solve for the hypotenuse. Well, you could say that a2+b2=c2, that's the Pythagorean theorem. And in this case, we said a and b are both five.

So we have 52+52=c2, and 52=25. So we have 25 + 25 equals c squared, 25 + 25 is 50. And what we can do is take the square root on both sides of this equation to get that c is equal to the square root of 50. And the square root of 50 actually simplifies down to five times the square root of two.

So notice when using the long version of this problem solving, we get to the same answer. But this is what's nice about the shortcut is it lets you get this answer without having to go through this long process. Now to ensure we know how to solve these types of triangles, let's see if we can solve some examples where we have this special case. So for each of these examples we're asked to solve for the unknown sides of each triangle. And we'll start with example a.

Now notice we have two 45-degree angles and two legs that are the same length. So that means we're dealing with a 45-45-90 triangle. Now recall that if we want to find the missing side or the hypotenuse, we just need to take one of the legs and multiply it by the square root of two. Well, one of the legs is 11 and then we multiply this by the square root of two. And that right there is the answer.

Notice how quick it is using this method. See, it's very straightforward, and that's what's really nice about these special cases. But now let's take a look at example b. In this example, we have a 45-degree angle, and we are given the hypotenuse. So how can we go about solving this?

Well, first off, we need to figure out if we actually are dealing with a special case triangle, and it turns out that we are. Because since we have a 45-degree angle here and a 90-degree cusp there, we know by default this has to be a 45-degree angle. You see, all the angles in a right triangle have to add up to 180, and 90 plus 45 plus 45 equals 180, so this is a special case triangle. Now to solve for the missing sides, what we can do is use this relationship. Notice in this situation, we're given the hypotenuse or the long side.

So what I'm going to do is take the hypotenuse instead of equal to the number we have, which is 13, and I'll say that that's equal to the leg multiplied by the square root of two. Now to solve for the leg, what I can do is divide the square root of two on both sides of this equation. That'll get the square roots to cancel, giving us that the leg of this triangle is equal to 13 over the square root of two. Now what I can do is rationalize the denominator here by multiplying the top and bottom by the square root of two. That'll get these square roots to cancel, giving us that the leg of this triangle is equal to 13 times the square root of two over two.

So what we're going to end up with is 13 radical two over two for this side of the triangle, and then 13 radical two over two for that side of the triangle. Because again, for a 45, 45, 90 triangle, these two sides have to have the same length. So that is how you can solve 45-45-90 triangles, and this is the shortcut that you can use. So hope you found this video helpful, thanks for watching, and let me know if you have any questions.

Given the triangle below, determine the missing side(s) without using the Pythagorean theorem (make sure your answer is fully simplified).

• A. $x=81$
• B. $x=9\sqrt2$
• C. $x=2\sqrt9$
• D. $x=\sqrt{162}$

Answer: (B)

Without using a calculator, determine all values of P in the interval $\left[0\degree,90\degree\right)$ with the following trigonometric function value.
$\csc P=\sqrt2$

• A. $P=30\degree$ only
• B. $P=45°$ only
• C. $P=60°$ only
• D. $P=30°,60°$

Answer: (B)

Common Trig Functions For 45-45-90 Triangles

Hey, everyone. So in a recent video, we talked about the special forty-five forty-five ninety right triangles, and we learned how you can use some shortcuts to quickly find the missing sides. Well, in this video, we're going to be learning about the common trig functions and how they are associated with the forty-five forty-five ninety triangle. And we're going to see if we can recognize any kinds of patterns for these trig functions because what we're going to learn for these special cases is there may be shortcuts when finding ratios for the trigonometric functions as well, meaning these triangles could be overall a lot easier to solve. So I'm always interested in finding shortcuts, and hopefully, you are too.

So let's get right into this. Now we're going to start with the sine function, and we know that sine is the opposite divided by the hypotenuse. Now if we go to a forty-five forty-five ninety right triangle, we can go to either one of the two angles because they're both 45 degrees. So if we take a look at this angle, for example, and we go to the opposite side, we can see that the opposite is five. And if we divide this by the hypotenuse or long side, that's going to be five times the square root of two.

We can cancel the fives here, giving us one over the square root of two. And by rationalizing this denominator, I'll multiply the top and bottom by the square root of two. This will get the square roots to cancel, giving us square root of two over two. So that means for the sine, we'll end up with square root of two over two. But now let's try solving for the cosine.

For the cosine, we get adjacent divided by hypotenuse. Now if we go to one of our 45 degree angles, the adjacent side is five. And if we divide this by the hypotenuse, we'll have five times the square root of two. The fives will cancel, giving us one over the square root of two. And we already figured out this is the same thing as square root of two over two, so this will just simplify to radical two over two, which is the cosine for this right triangle.

Now let's take a look at the tangent. For the tangent, we have opposite over adjacent. So what I'm going to do is go over here to our right triangle, look at one of our angles, and if I go to the opposite side, we end up with five. If I go to the adjacent side, we also have five. So we'll have five over five, which is simply one, meaning the tangent for our forty-five forty-five ninety triangle is one.

So let's see if we notice any patterns with these trigonometric functions. Well, something that I notice is that the sine and cosine are the same, and the tangent just comes out to one. So this is something that's pretty straightforward about these triangles is that you get the same sine and cosine values. But let's take a look at these other reciprocal identities like the cosecant, secant, and cotangent. So for the cosecant, we know that this is just one over the sine of theta, which means all we're going to do is flip this fraction that we got.

So we'll have two over the square root of two, and then we can rationalize this denominator. When doing this, the square roots will cancel at the bottom of the fraction, giving us two times the square root of two over two. These are going to cancel right here, meaning all we're going to end up with is the square root of two for the cosecant. And because the sine and cosine were the same, that means one over the cosine, which is our secant, is also going to be the square root of two. So all we now have to solve for is the cotangent, which is just one over the tangent, but we learned that the tangent is one.

So the cotangent will be one over one, which is simply equal to one. So notice how really straightforward all these trigonometric functions are when you're looking at a special case forty-five forty-five ninety triangle. We can recognize that the sine and cosine will always be square root of two over two, that the cosecant and secant will always just be the square root of two, and that the tangent and cotangent will always be one. So these are all of the common trig functions for the forty-five forty-five ninety triangle. Hope you found this video helpful, and thanks for watching.

30-60-90 Triangles

Welcome back, everyone. So in recent videos, we've been talking about special case right triangles. And we recently learned about the 45-45-90 triangle and some shortcuts you can use for solving this special case. Well, in this video, we're going to be taking a look at the 30-60-90 triangle, which is another special case. And for these triangles, there are shortcuts you can use to solve them as well.

This type of triangle is going to show up a lot, not only in this course but also in future math courses and possibly other courses too, like physics or other science-related courses, because this situation just tends to happen a lot. So without further ado, let's get right into some shortcuts you can use to solve this triangle. The thing that's special about this triangle is that you can relate the side lengths to the shortest leg of the triangle. Keep in mind for the 30-60-90 triangle, it's a bit different than the 45-45-90 triangle. Because for the 45-45-90 case, we learned that two of the side lengths are the same for the triangle, and so you can use one relationship to relate all the sides.

For the 30-60-90 case, the two side lengths are not the same. We have two different lengths. So what we need to do here is keep track of which side relates to which other side, but we can do that using the shortcuts. For example, if you want to find the hypotenuse of a 30-60-90 triangle, you could take the short leg of your triangle, the shortest side, and you can multiply it by two. And if you want to find the long leg of your triangle, you can take the short leg of the triangle again, and you can multiply it by the square root of three.

Now, again, I want to emphasize, these shortcuts only work if you're dealing with a 30-60-90 triangle, and this is the special case we have in this example. Let's see if we can solve for the missing sides. I can see that the hypotenuse is going to be the shortest leg multiplied by two. I can see the shortest leg is five. And if we multiply that by two, we're going to get 10, meaning the hypotenuse is 10.

Now I can see that the long leg of the triangle is equal to the short leg, which we have down here, and we said that that's five, multiplied by the square root of three. So the long leg of the triangle is going to be five×3. So this is how you can find the missing sides of a 30-60-90 right triangle. As you can see, it's still very straightforward when you have the shortcuts available. But to make sure we know how to use these shortcuts well, let's try some other examples that are a little bit more complicated.

For these examples, we're asked to solve for the unknown side of each triangle, and we're going to start with example a. In this example, I see that we have the hypotenuse and a 30-degree angle. So what we first have to ask ourselves is, is this a 30-60-90 triangle? Well, it actually is. Because if this is 30 degrees and this is 90 degrees, then the other missing angle has to be 60 degrees.

Because 60 plus 30 gives you 90, and then you add another 90, you get to one eighty. And all angles in the triangle have to add to one eighty degrees. So this is a special case. So that means we can use the shortcuts we learned about. First, I'm going to try solving for the short leg, and I'm going to do this using this relationship up here.

Because we have that the hypotenuse is equal to the short leg multiplied by two. Now the hypotenuse, we can see, is the longest side, which is eight, and so that's going to equal the short leg of the triangle multiplied by two. So if I want to find the short leg, all I need to do is divide both sides of this equation by two. That's going to get the twos to cancel on the right side, giving us that the short leg of our triangle is equal to eight divided by two, which is four. So that means this missing side is four.

If I want to find the long side of the triangle, I could use the Pythagorean theorem from here, or I could use this relationship that says that the long leg is the short leg multiplied by the square root of three. So we have that the long leg is equal to the four×3. Meaning this missing side of the triangle is four×3. So that is how you can use shortcuts to find the missing sides of a 30-60-90 triangle.

Now let's try another example, in this case, example b. And for this example, we're asked to find the missing sides as well. And what I can see is that we have a 60-degree angle and the long side of the triangle given to us. Now because we have a 60-degree angle here, by default, this other angle has to be 30 degrees, because all angles in the triangle have to add to one eighty. So this is a special case.

Since I see that we have the long leg, I'm going to use this relationship right here, which says that the long leg is going to be equal to the short leg multiplied by the square root of three. Now I can see that we have that the long leg of the triangle is one, so we'll have one, the long side, is equal to the short side multiplied by the square root of three. So what I can do from here is divide the square root of three on both sides of this equation, which will give the square root of three's to cancel on the right side, giving us that the short leg of this triangle is equal to one divided by the square root of three. And I can go ahead and rationalize this denominator by multiplying the top and bottom by the square root of three.

That'll get the square roots to cancel, giving us that the short leg of this triangle is equal to the 3÷3. So that's going to be the short leg of the right triangle. Our last step for solving this right triangle is to find the hypotenuse or the longest side. But I can see based on the relationship up here that the hypotenuse is going to be the short leg, which we already determined is 3÷3 multiplied by two. So that means that this missing side of the triangle is going to be two times the 3÷3.

So this is how you can find all the missing sides of a 30-60-90 triangle using some simple shortcuts. That's how you solve these types of problems. I hope you found this video helpful. Thanks for watching.

Given the triangle below, determine the missing side(s) without using the Pythagorean theorem (make sure your answer is fully simplified).

• A. $x=10,y=5\sqrt5$
• B. $x=3,y=4$
• C. $x=5\sqrt3,y=10$
• D. $x=5,y=5\sqrt2$

Answer: (C)

Without using a calculator, determine all values of A in the interval $\left[0,\frac{\pi}{2}\right)$ with the following trigonometric function value.
$\cos A=\frac{\sqrt3}{2}$

• A. $0$ only
• B. $\frac{\pi}{4}$ only
• C. $\frac{\pi}{6}$ only
• D. $\frac{\pi}{3}$ only

Answer: (C)

Common Trig Functions For 30-60-90 Triangles

Welcome back, everyone. So up to this point, we've been talking about special case right triangles, and we recently took a look at the thirty, sixty, 90 case. Well, in this video, we're going to be taking a look at the trigonometric functions associated with the thirty, sixty, 90 triangle. So far, the trigonometric functions we've learned about have been a bit tedious just because we have to find ratios for each of the situations and see how they relate to each other. But what we're hopefully going to figure out in this video is that for these special case triangles, there are patterns that show up when you solve for the trigonometric ratios and functions.

So without further ado, let's get right into this and see if we can find some shortcuts and patterns that show up. So we're going to start with finding the sine of a thirty sixty, 90 triangle, and we're going to focus on the 30 degree angle for these trigonometric functions on the left side. Recall that sine is opposite over hypotenuse. So if I go to our 30 degree angle, the opposite side of this triangle is five, and the hypotenuse, or the long side of the triangle, is 10. And five over 10 reduces to one two, meaning that the sine of 30 degrees is one two.

But now let's take a look at the cosine of our 30 degree angle. For the cosine, it's going to be adjacent over hypotenuse. And the adjacent side to the 30 degree angle is five times the square root of three, and then this is divided by the hypotenuse, which is 10. And again, five over ten reduces to one half, so we'll end up with one times the square root of three over two. And one times the square root of three is just the square root of three, so all we're going to end up with is the square root of three over two.

And that's our cosine. But now let's take a look at the tangent. For the tangent, it's going to be opposite over adjacent. Well, going to this triangle, we can see that the opposite side is five, and then the adjacent side is five times the square root of three. Again, the fives will cancel giving us one over the square root of three.

And if I go ahead and rationalize this denominator by multiplying the top and bottom by the square root of three, we'll get the square roots to cancel, giving us the square root of three over three as our tangent. So this is what we get for the sine, cosine, and tangent of our thirty, sixty, 90 triangle when specifically looking at the 30 degree angle. But now let's solve for the cosecant, and we're going to use reciprocal identities to solve these next three trigonometric functions. Because I noticed for the cosecant, it's the same thing as one over the sine of theta. And we said that the sine of theta is one half, so all we need to do is flip one half, which is just going to be two over one or two.

So the cosecant of our 30 degree angle is two. Now let's take a look at the secant of our angle. Well, for the secant, we can see that this is one over cosine, and we already said that the cosine is square root three over two. So for secant, we're going to get two over the square root of three, which I will need to rationalize this denominator. So multiply the top and bottom by the square root of three, cancel the square roots, giving us two times the square root of three over three, which is the result for our secant of 30 degrees.

But now let's take a look at our cotangent. For the cotangent, it's going to be one over the tangent. And we figured out that the tangent is square root three over three. So the cotangent is going to be three over the square root of three, just flipping this fraction. And again, I'll need to rationalize this denominator.

When doing this, the square roots will cancel, giving us three times the square root of three over three. And these threes are going to cancel, leaving us with just the square root of three. So that is how you can find the trigonometric functions for the 30 degree angle. Now because this is a thirty, sixty, 90 triangle, we also need to find all the trigonometric functions for the 60-degree angle. So let's go ahead and do that on this right side.

So for the sine function, if we're looking for 60 degrees, recall that this is opposite over hypotenuse. Now the opposite side is five times the square root of three, and this is going to be divided by the hypotenuse, which is 10. The five and ten will reduce to give us just a two in the denominator, meaning that all we're going to have is the square root of three over two for our sine. Now let's take a look at our cosine. For the cosine of 60 degrees, recall that this is adjacent over hypotenuse, so we're going to go to the adjacent side of the 60-degree angle, which is five, and then divide this by the hypotenuse, which is 10.

Five over 10 reduces to one half, meaning that the cosine over 60-degree angle is one half. Now let's take a look at the tangent, which is opposite over adjacent. The side opposite to 60 degrees is five times the square root of three, and then the adjacent side is five. The fives will cancel giving us just the square root of three, meaning that the square root of...

Here’s what students ask on this topic:

What is SOHCAHTOA and how does it help in solving trigonometric problems?

SOHCAHTOA is a mnemonic that helps remember the relationships between angles and side lengths in right triangles for trigonometric functions. It stands for:

$\sin \left(\theta \right)=\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}$

$\cos \left(\theta \right)=\frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$

$\tan \left(\theta \right)=\frac{\mathrm{opposite}}{\mathrm{adjacent}}$

It simplifies solving problems by providing a clear formula for each function. For example, to find the sine of an angle, you divide the length of the opposite side by the hypotenuse. This tool is especially useful for memorizing and applying trigonometric ratios efficiently.

How do you calculate the sine, cosine, and tangent of an angle in a right triangle?

To calculate the sine, cosine, and tangent of an angle in a right triangle, use the following formulas:

$\sin \left(\theta \right)=\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}$

$\cos \left(\theta \right)=\frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$

$\tan \left(\theta \right)=\frac{\mathrm{opposite}}{\mathrm{adjacent}}$

Identify the angle you are working with, then determine the opposite, adjacent, and hypotenuse sides relative to that angle. Plug the side lengths into the respective formula. For example, if the opposite side is 3 and the hypotenuse is 5, the sine of the angle is $\frac{3}{5}$.

What are reciprocal identities in trigonometry?

Reciprocal identities in trigonometry are relationships between primary trigonometric functions and their reciprocals. They are defined as:

$\csc \left(\theta \right)=\frac{1}{\sin }$

$\sec \left(\theta \right)=\frac{1}{\cos }$

$\cot \left(\theta \right)=\frac{1}{\tan }$

For example, if $\sin \left(\theta \right)=\frac{3}{5}$, then $\csc \left(\theta \right)=\frac{5}{3}$. Reciprocal identities simplify calculations and are essential for solving complex trigonometric problems.

What are the shortcuts for solving 45-45-90 triangles?

In a 45-45-90 triangle, the two legs are equal, and the hypotenuse is related to the leg length by a simple ratio. The shortcuts are:

$\mathrm{hypotenuse}=\mathrm{leg}\times \sqrt{2}$

For example, if the leg length is 5, the hypotenuse is $5\times \sqrt{2}$, which simplifies to $5\sqrt{2}$. This shortcut eliminates the need for the Pythagorean theorem, making calculations faster and more efficient.

How do you find trigonometric functions for a 30-60-90 triangle?

In a 30-60-90 triangle, the side lengths follow specific ratios: the shortest leg is opposite the 30° angle, the hypotenuse is twice the shortest leg, and the longer leg is the shortest leg multiplied by the square root of three. To find trigonometric functions:

$\sin \left(30\right)=\frac{1}{2}$

$\cos \left(30\right)=\frac{\sqrt{3}}{2}$

$\tan \left(30\right)=\frac{1}{\sqrt{3}}$

For the 60° angle, the sine and cosine values switch, and the tangent becomes $\sqrt{3}$. These patterns simplify calculations for this special triangle.