Class 10 math (India)
- Special right triangles intro (part 1)
- Special right triangles intro (part 2)
- Trigonometric ratios of special angles
- 30-60-90 triangle example problem
- Special right triangles
- Special right triangles proof (part 1)
- Special right triangles proof (part 2)
- Evaluating expressions of trigonometric ratios for some special angles
Special right triangles proof (part 1)
Learn how to prove the ratios between the sides of a 30-60-90 triangle. Created by Sal Khan.
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- this is a little confusing, especially at the beginning. help(25 votes)
- I would like to help, but please be more specific on what is confusing to you.(6 votes)
- At6:20, Sal mentions taking the "principal root" of something. What is a principal root, and how is it different than square root? Thanks!(7 votes)
- The square root of a number is both positive and negative:
√9 = ±3
Because both 3 and -3 squared give you 9. The principal root is just the positive answer. So the principal square root of 9 is 3.(21 votes)
- how did u get x squared over four?(15 votes)
- (x/2) squared means you square both the numerator and the denominator, so that's where the x^2 over 4 comes from.(8 votes)
- When writing out the Pythagorean theorem, I'm a bit confused about why I'm not allowed to take the square root of each term.
So, a^2 + b^2 = c^2
Take the root of both sides
a + b = c(3 votes)
- It is quite tempting to think that the square root of (a^2 + b^2) is a+b, but this simply does not work. The easiest way to see this is to plug in numbers for a and b, such as a=3 and b=4.
sqrt(3^2+4^2) = sqrt(9+16) = sqrt(25) = 5.
However, 3+4 = 7, which is not 5.(7 votes)
- I know that an
~above it means congruent, but what does just a
~mean in math? Or does it even have to do with math? And if it is used in math, what would you use it for, and when does it come in, in math (like which grade?).(4 votes)
- Often, ~ is used to mean "approximately equal to".
e.g. 1.7498 ~ 1.75
It would typically be introduced around the same time you learn about decimals and fractions (your school may vary).(4 votes)
- 5:54: where does that 4 come from?(4 votes)
- That's because (x/2)^2 = (x^2)/(2^2) = (x^2)/4. When you square a fraction, you square both the numerator and the denominator. Pay attention to parentheses, and you'll be fine.
If you're asking about that 4 he glued to x^2 on the right side, then don't forget that he immediately divided 4x^2 by 4, which is the same as x^2 since 4/4 = 1. This is a useful trick to help you add/subtract fractions without multiplying everything (in both sides) by what's in the denominator in order to get rid of fractions. He'd have to divide by 4 again, anyway, in order to solve for BD.(4 votes)
- could you have a right triangle which has the following: a right angle, an angle of 89 and an angle of 1(3 votes)
- Yep! Just make sure that the angles all add up to 180º. Hope this helps!
- Shouldn't the side opposite the 60 degrees angle be double the side opposite the 30 degrees angle? Logically, wouldn't it make sense that an angle that is double the size of another angle, should have a corresponding segment that is double the segment made by the smaller angle? Why is this not the case?(3 votes)
- You are suggesting that the sides should be in proportion to the angles but in fact there is no reason this should be true.
Here is one argument against it.
If the sides were in proportion to the angles, then the hypotenuse (the side opposite the 90 degree angle) would be triple the side opposite the 30 degree angle. The sides would be 1, 2, 3 or 2, 4, 6, etc. This is clearly impossible since the third side has to be shorter than the sum of the other 2 sides, since the shortest side is a straight line. Another reason we know this isn't true is because it wouldn't satisfy the pythagorean theorem 1^2 + 2^2 doesn't = 3^2.
Hope this helps.(3 votes)
- Is there any way to print out the transcript of your presentations. Having terrible time trying to remember everything you say.(3 votes)
- Try taking notes on paper about the transcript along with the video so you remember how to do the problems and can refer back to it more easily when practicing the concepts(3 votes)
- @2:10is "dropping an altitude" the same as bisecting the top angle? Do both angle bisector and an altitude split the bottom side at the midpoint?(2 votes)
- No, an altitude does not necessarily bisect the angle. The only cases where an altitude bisects the angle from which it is dropped is in an equilateral triangle, or in an isosceles triangle when you drop an altitude from the angle that's formed by the two equal sides (doesn't work with the two other angles). But you don't need to memorize that, it's just as an FYI!
As for your second question: no, neither of them necessarily bisect the opposite side at the midpoint. Only perpendicular bisectors and medians (as a reminder, medians are the line segments that go from the vertex of an angle to the midpoint of the opposite side; the three medians intersect at the centroid) always bisect the opposite side. BUT, like in the above question, an altitude and an angle bisector CAN bisect the opposite side, but they're special cases.(4 votes)
What I want to do in this video is discuss a special class of triangles called 30-60-90 triangles. And I think you know why they're called this. The measures of its angles are 30 degrees, 60 degrees, and 90 degrees. And what we're going to prove in this video, and this tends to be a very useful result, at least for a lot of what you see in a geometry class and then later on in trigonometry class, is the ratios between the sides of a 30-60-90 triangle. Remember, the hypotenuse is opposite the 90-degree side. If the hypotenuse has length x, what we're going to prove is that the shortest side, which is opposite the 30-degree side, has length x/2, and that the 60 degree side, or the side that's opposite the 60-degree angle, I should say, is going to be square root of 3 times the shortest side. So square root of 3 times x/2, that's going to be its length. So that's where we're going to prove in this video. And then in other videos, we're just going to apply this. We're going to show that this is actually a pretty useful result. Now, let's start with a triangle that we're very familiar with. So let me draw ourselves an equilateral triangle. So drawing the triangles is always the hard part. This is my best shot at a equilateral triangle. So let's call this ABC. I'm just going to assume that I've constructed an equilateral triangle. So triangle ABC is equilateral. And if it's equilateral, that means all of its sides are equal. And let's say equilateral with sides of length x. So this is going to be x, this is going to be x, and this is going to be x. We also know, based on what we've seen from equilateral triangles before, that the measures of all of these angles are going to be 60 degrees. So this is going to be 60 degrees, this is going to be 60 degrees, and then this is going to be 60 degrees. Now, what I'm going to do is I'm going to drop an altitude from this top point right over here. So I'm going to drop an altitude right down, and by definition, when I'm constructing an altitude, it's going to intersect the base right here at a right angle. So that's going to be a right angle, and then this is going to be a right angle. And it's a pretty straightforward proof to show that not only is this an altitude, not only is it perpendicular to this base, but it's a pretty straightforward proof to show that it bisects the base. And you could pause it, if you like and prove it yourself. But it really comes out of the fact that it's easy to prove that these two triangles are congruent. So let me prove it for you. So let's call this point D right over here. So triangles ABD and BDC, they clearly both share this side. So this side is common to both of them right over here. This angle right over here is congruent to this angle over there. This angle right over here is congruent to this angle over here. And so if these two are congruent to each other, then the third angle has to be congruent to each other. So this angle right over here needs to be congruent to that angle right over there. So these two are congruent. And so you can use actually a variety of our congruence postulates. We could say, side-angle-side congruence. We could use angle-side-angle, any of those to show that triangle ABD is congruent to triangle CBD. And what that does for us, and we could use, as I said, we could use angle-side-angle or side-angle-side, whatever we like to use for this. What that does for us is it tells us that the corresponding sides of these triangles are going to be equal. In particular, AD is going to be equal to CD. These are corresponding sides. So these are going to be equal to each other. And if we know that they're equal to each other, and they add up to x-- remember, this was an equilateral triangle of length x-- we know that this side right over here, is going to be x/2. We know this is going to be x/2. Not only do we know that, but we also knew when we dropped this altitude, we showed that this angle has to be congruent to that angle, and their measures have to add up to 60. So if two things are the same and they add up to 60, this is going to be 30 degrees, and this is going to be 30 degrees. So we've already shown one of the interesting parts of a 30-60-90 triangle, that if the hypotenuse-- notice, and I guess I didn't point this out. By dropping this altitude, I've essentially split this equilateral triangle into two 30-60-90 triangles. And so we've already shown that if the side opposite the 90-degree side is x, that the side opposite the 30-degree side is going to be x/2. That's what we showed right over here. Now we just have to come up with the third side, the side that is opposite the 60-degree side. I'll just use the letters that we already have here. This is BD. And we can just use the Pythagorean theorem right here. BD squared plus this length right over here squared plus x/2 squared is going to be equal to the hypotenuse squared. So we get BD squared plus x/2 squared-- this is just straight out of the Pythagorean theorem.-- plus x/2 squared is going to equal this hypotenuse squared. It's going to equal x squared. And just to be clear, I'm looking at this triangle right here. I'm looking at this triangle right over here on the right, and I'm just applying the Pythagorean theorem. This side squared plus this side squared is going to equal the hypotenuse squared. And let's solve now for BD. You get BD squared plus x squared over 4. x squared over 4 is equal to x squared. You could view this as 4x squared over 4. That's the same thing, obviously, as x squared. If you subtract 1/4 x squared from both sides, or x squared over 4 from both sides, you get BD squared is equal to-- 4x squared over 4 minus x squared over 4 is going to be 3x squared over 4. So it's just going to be 3x squared over 4. Take the principal root of both sides. You get BD is equal to the square root of 3 times x. The principal root of 3 is square root of 3. the principal root of x squared is just x, over the principal root of 4 which is 2. And BD is the side opposite the 60-degree side. So we're done. If this hypotenuse is x, the side opposite the 30-degree side is going to be x/2, and the side opposite the 60-degree side is going to be square root of 3 over 2 times x, or the square root of 3x over 2, depending on how you want to view it.