Special Right Triangles - Online Tutor, Practice Problems & Exam Prep

Special right triangles, such as the 45-45-90 and 30-60-90 triangles, offer shortcuts for solving for unknown sides. In a 45-45-90 triangle, the hypotenuse is found by multiplying a leg by $\sqrt{2}$. For a 30-60-90 triangle, the hypotenuse is double the shortest leg, while the long leg is the short leg multiplied by $\sqrt{3}$. Trigonometric functions reveal patterns, with sine and cosine values switching between angles, enhancing problem-solving efficiency.

45-45-90 Triangles

Welcome back, everyone. So, up to this point, we have 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. Specifically, we're going to be talking about the 45-45-90 special triangle. The reason these triangles are special is 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. 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. 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 2. Because if you take a leg, like 5, and you multiply it by the square root of 2, 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 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 is equal to c, and we want to solve for the hypotenuse. Well, you could say that \( a^2 + b^2 = c^2 \), that's the Pythagorean theorem. And in this case, we said a and b are both 5. So we have \( 5^2 + 5^2 = c^2 \), and \( 5^2 \) is 25. So we have 25 + 25 = \( c^2 \). 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 5 times the square root of 2. 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; 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 2. Well, one of the legs is 11, and then we multiply this by the square root of 2. 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 are actually 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 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 I'm going to do is take the hypotenuse, set it 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 2. Now to solve for the leg, what I can do is divide the square root of 2 on both sides of this equation. That'll get the square root of twos to cancel, giving us that the leg of this triangle is equal to 13 over the square root of 2. Now what I can do is rationalize the denominator here by multiplying the top and bottom by the square root of 2. 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 2 over 2. So what we're going to end up with is 13 radical 2 over 2 for this side of the triangle and then 13 radical 2 over 2 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 I 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. $x=\sqrt{162}$

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. $P=30°,60°$

Common Trig Functions For 45-45-90 Triangles

Hey, everyone. So in a recent video, we talked about the special 45, 45, 90 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 45, 45, 90 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 45-45-90 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 5. And if we divide this by the hypotenuse or long side, that's going to be 5 times the square root of 2. We can cancel the fives here, giving us 1 over the square root of 2. And by rationalizing this denominator, I'll multiply the top and bottom by the square root of 2. This will get the square roots to cancel, giving us square root of 2 over 2.

So that means for the sine, we'll end up with 22. 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 5. And if we divide this by the hypotenuse, we'll have 5 times the square root of 2. The fives will cancel, giving us 1 over the square root of 2. We already figured out this is the same thing as square root of 2 over 2, so this will just simplify to radical 2 over 2, 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 gonna 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 5. If I go to the adjacent side, we also have 5. So we'll have 5 over 5, which is simply 1, meaning the tangent for our 45, 45, 90 triangle is 1.

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 1. 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 gonna do is flip this fraction that we got. So we'll have 2 over the square root of 2, and then we can rationalize this denominator. When doing this, the square roots will cancel on the bottom of the fraction, giving us 2 times the square root of 2 over 2. These are going to cancel right here, meaning all we're going to end up with is the square root of 2 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 2.

So all we now have to solve for is the co-tangent, which is just one over the tangent, but we learned that the tangent is 1. So the cotangent will be 1 over 1, which is simply equal to 1. So notice how really straightforward all these trigonometric functions are when you're looking at a special case 45, 45, 90 triangle. We can recognize that the sine and cosine will always be square root 2 over 2, that the cosecant and secant will always just be the square root of 2, and that the tangent and co-tangent will always be 1. So these are all of the common trig functions for the 45, 45, 90 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. And this is a type of triangle that 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.

Now the thing that's special about this triangle is that you can relate the side lengths to the shortest leg on the triangle. Now 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. Well, 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. So, for example, if you want to find the hypotenuse of a 30-60-90 triangle, you can take the short leg of your triangle, the shortest side, and you can multiply it by 2. 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 3. 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. So let's see if we can solve for the missing sides.

Well, I can see that the hypotenuse is going to be the shortest leg multiplied by 2, I can see the shortest leg is 5, and if we multiply that by 2 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 5 multiplied by the square root of 3. So the long leg of the triangle is going to be 5 × 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.

So for these examples, we're asked to solve for the unknown side of each triangle, and we're going to start with example a. Now 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 180. And all angles in the triangle have to add to 180 degrees. So this is a special case. So that means we can use the shortcuts we learned about. Well, first, I'm gonna 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 2. Now the hypotenuse, we can see, is the longest side, which is 8, and so that's going to equal the short leg of the triangle multiplied by 2. So if I want to find the short leg, all I need to do is divide both sides of this equation by 2. 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 8 divided by 2, which is 4. So that means this missing side is 4. Now 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 3. So we have that the long leg is equal to the short leg, which we just figured out is 4, and that's multiplied by the square root of 3. Meaning, this missing side of the triangle is 4 times the square root of 3.IRONMENT_KEYS_SUBMIT_CAPTION_SINGULAR}

So that is how you can use shortcuts to find the missing sides of a 30-60-90 triangle. But now let's try another example, in this case, example b.

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. $x=5,y=5\sqrt2$

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

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 30, 60, 90 case. Well, in this video, we're going to be taking a look at the trigonometric functions associated with the 30, 60, 90 triangle. Now 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 function. 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 30, 60, 90 triangle, and we're going to focus on the 30 degree angle for these trigonometric functions on the left side. Now recall that sine is opposite over hypotenuse. So if I go to our 30-degree angle, the opposite side of this triangle is 5, and the hypotenuse or the long side of the triangle is 10, and 510 reduces to 12, meaning that the sine of 30 degrees is 1/2.

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 5 times the square root of 3, and then this is divided by the hypotenuse which is 10. And again, 510 reduces to one-half, so we'll end up with 1 times the square root of 3 over 2, and one times the square root of 3 is just the square root of 3, so all we're gonna end up with is the square root of 3 over 2, 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 5, and then the adjacent side is 5 times the square root of 3. Again, the fives will cancel giving us 13, and if I go ahead and rationalize this denominator by multiplying the top and bottom by the square root of 3, we'll get the square roots to cancel, giving us the square root of 3 over 3 as our tangent.

So this is what we get for the sine, cosine, and tangent of our 30, 60, 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 3 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 1/2. So all we need to do is flip 1/2, which is just going to be 2 over 1 or 2. So the cosecant of our 30-degree angle is 2.

Now let's take a look at the secant of our angle. Well, for the secant, we can see that this is 1 over cosine, and we already said that the cosine is square root 3 over 2. So for secant, we're going to get 23, which I will need to rationalize this denominator, so multiply the top and bottom by the square root of 3, cancel the square roots giving us 2 times the square root of 3 over 3, 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 1 over the tangent. We figured out that the tangent is square root 3 over 3. So the cotangent is going to be 3 over the square root of 3, just flipping this fraction. And again, I'll need to rationalize this denominator. When doing this, the square roots will cancel, giving us 3 times the square root of 3 over 3, and these threes are going to cancel leaving us with just the square root of 3.

So that is how you can find the trigonometric functions for the 30-degree angle. Now because this is a 30, 60, 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 5 times the square root of 3, and this is going to be divided by the hypotenuse, which is 10. The 5/10 will reduce to give us just a 2 in the denominator, meaning that all we're going to have is the square root of 3 over 2 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 5, and then divide this by the hypotenuse which is 10. 510 reduces to 1 half, meaning that the cosine of the 60-degree angle is 1/2.

Now let's take a look at the tangent, which is opposite over adjacent. The side opposite to 60 degrees is 5 times the square root of 3, and then the adjacent side is 5. The fives will cancel giving us just the square root of 3, meaning that the square root of 3 is the tangent of 60 degrees. Now for these last 3 trigonometric functions, we're going to use the reciprocal identities. Now cosecant is the same thing as one over sine. So if I want to find the cosecant of 60 degrees, I just need to flip the sine of 60 degrees, which would be 23.

Now I can go ahead and rationalize this denominator. That'll get the square roots to cancel on the bottom, giving us 2 times square root 3 over 3. So that is going to be the cosecant of our 60-degree angle. Now what I can also do is find the secant of our 60-degree angle. And to do this, what I need to do is it's one over cosine, so I need to go ahead and flip the cosine of 60 degrees. Since the cosine of 60 degrees is 1/2, that means that the secant of 60 degrees is going to be 2 over 1, which is just equal to 2. So 2 is the secant of 60 degrees.

Now lastly, we're gonna take a look at the cotangent of 60 degrees, and all I need to do is go ahead and flip the tangent that we got here. So we said the tangent is square root of 3, so the cotangent is gonna be 13. I can go ahead and rationalize the denominator, getting the square roots to cancel on the bottom, giving us the square root of 3 over 3 for the cotangent of 60 degrees.

So these are all of the trigonometric functions evaluated for the 30, 60, 90 triangle. And you may notice that we have some patterns that emerge when we figure out what all these trigonometric values are, because notice how the sine and cosine will switch places when looking at the different angles. Because the sine of 30 is 1/2, whereas the cosine of 30 is square root 3 over 2. But when we go to the 60-degree angle, the sine is square root 3 over 2, and the cosine is 1/2. We see the same kind of behavior in the cosecant and secant. Since these are reciprocal identities, you just switch them when looking at the different angles.

Now we also see this behavior in the tangent and cotangent. Because notice when looking at the 30-degree angle, the tangent is square root 3 over 3, and the cotangent is square root 3. But when looking at the 60-degree angle, the tangent is just square root 3, and the cotangent is square root 3 over 3. So we see the same kind of switch when looking at this trigonometric function.

So hopefully this helped you to recognize some of the patterns and give you some general understanding as to how you can find the trigonometric functions for a 30, 60, 90 triangle. Hope you found this video helpful. Thanks for watching.

Here’s what students ask on this topic:

What are the side length relationships in a 45-45-90 triangle?

In a 45-45-90 triangle, the two legs are of equal length. If each leg is denoted as a, then the hypotenuse is a√2. This relationship can be derived from the Pythagorean theorem, but it serves as a useful shortcut for quickly solving these triangles. For example, if each leg is 5 units, the hypotenuse would be 5√2 units.

How do you find the missing sides of a 30-60-90 triangle?

In a 30-60-90 triangle, the side lengths have specific relationships. The hypotenuse is twice the length of the shortest leg, and the longer leg is the shortest leg multiplied by √3. If the shortest leg is a, then the hypotenuse is 2a and the longer leg is a√3. For example, if the shortest leg is 4 units, the hypotenuse is 8 units, and the longer leg is 4√3 units.

What are the trigonometric functions for a 45-45-90 triangle?

For a 45-45-90 triangle, the trigonometric functions are consistent due to the equal angles. The sine and cosine of 45 degrees are both √2/2. The tangent of 45 degrees is 1. The reciprocal functions are: cosecant (csc) is √2, secant (sec) is √2, and cotangent (cot) is 1. These values simplify solving trigonometric problems involving 45-45-90 triangles.

What are the trigonometric functions for a 30-60-90 triangle?

In a 30-60-90 triangle, the trigonometric functions vary based on the angle. For the 30-degree angle, sine is 1/2, cosine is √3/2, and tangent is √3/3. For the 60-degree angle, sine is √3/2, cosine is 1/2, and tangent is √3. The reciprocal functions follow: cosecant (csc) of 30 degrees is 2, secant (sec) of 30 degrees is 2/√3, and cotangent (cot) of 30 degrees is √3. These patterns help in quickly solving trigonometric problems.

How do you solve for the hypotenuse in a 45-45-90 triangle?

To solve for the hypotenuse in a 45-45-90 triangle, you can use the relationship that the hypotenuse is the leg length multiplied by √2. If each leg is a, then the hypotenuse c is given by c = a√2. For example, if each leg is 7 units, the hypotenuse would be 7√2 units.