WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi, these are the trigonometry lectures for educator.com.
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Today, we're going to talk about trigonometry in triangles that have a right angle.
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These are called right triangles.
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The master formula for right triangles, we've seen it before it's the SOH CAH TOA.
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That's the word that I remember to know that the sin(θ) ...
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Let me draw a right triangle here.
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If θ is one of the small angles, not the right angle, then the sin(θ) is equal to the length of the opposite side over the length of the hypotenuse.
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Cos(θ) is equal to the adjacent side over the hypotenuse.
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The tan(θ) is equal to the opposite side over the adjacent side.
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For shorthand, sine is equal to opposite over hypotenuse, cosine is equal to adjacent over hypotenuse, tangent is equal to opposite over adjacent.
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It's probably worth saying SOH CAH TOA often enough until it sticks into your memory, because it really is useful for remembering these things.
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If you have a hard time remembering that, the little mnemonic that is also helpful for some students is Some Old Horse Caught Another Horse Taking Oats Away.
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That spells out SOH CAH TOA for you.
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One key thing to remember here is that SOH CAH TOA only works in right triangles.
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You have to have one angle being a right angle.
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If you have a triangle that is not a right triangle, if you don't have a right angle, don't use SOH CAH TOA because it's not valid in triangles that don't have right angles.
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That's if you have no right angle.
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We're going to learn in the next lectures on educator.com, we'll learn about the law of sines and the law of cosines.
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Those work in any triangle where you don't need a right angle, but when you have a right angle, it's definitely easier and better and quicker to use SOH CAH TOA.
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Let's try that out with some actual triangles.
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On the first example, we have a right triangle with short sides of length 3 and 4, and we want to find all the angles in the triangle.
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Let me draw a triangle here.
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We're told that the short sides have length 3 and 4.
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Of course, one angle is a right angle, so I don't need to worry about that.
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I'll call these angles θ, and I'll call this one φ.
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First thing we're going to need to know is what the hypotenuse of this triangle is.
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h²=3²+4², Pythagorean theorem there, which is 9+16, which is 25.
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So, h is the square root of 25, h is 5.
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Let me draw that in there.
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Now, I want to figure out what θ and φ are.
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I'm going to use SOH CAH TOA.
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Let me write that down there for reference, SOH CAH TOA.
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I'm going to figure out what θ is by using the SOH part of the SOH CAH TOA.
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Sin(θ) is equal to the opposite over the hypotenuse.
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The opposite angle to θ is 3 and the hypotenuse is 5.
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So, θ=arcsin(3/5).
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I'm going to work that out on the calculator.
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My calculator has an arcsine button, it actually writes it as sin^-1, which I don't like that notation because makes it seem like a power.
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In any case, I'm going to use inverse sine of 3/5.
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There's a very important step here that many students get confused about which is that, if you're looking for an answer in terms of degrees, which in real world measurement, it sometimes easier to use degrees than radians.
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You have to set your calculator to degree mode.
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Most calculators have a degree mode and a radian mode.
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In fact, all calculators that do trigonometric functions have a degree mode and a radian mode.
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The default is probably radian mode.
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If your calculator is set in radian mode and you try to do something like arcsin(3/5), you'll get an answer that doesn't look right.
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You may be confused if you're checking your answers somewhere, it may not agree with what the correct answer it.
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What you have to do is set up your calculator in degree mode, if you want an answer in degrees.
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My calculator is a Texas Instruments.
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It's got a mode button, I just scroll down, it say's RAD and DEG to convert it from radians to degrees.
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I'm going to convert it into degree mode and then I'll get an answer in terms of degrees.
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That's a step that many students forget and they get kind of confused when they get an answer which is in terms of radians, but it doesn't agree with the degree answer they were looking for.
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Now, I've got my calculator set in degree mode.
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I'll do the arcsin(3/5) which is 0.6.
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It tells me that that is approximately equal to 36.9 degrees.
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I found one of the angles in the triangle.
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Of course, another one is a right angle, so it's 90 degrees.
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For φ, I think for φ, I'm going to practice the cosine part of the SOH CAH TOA.
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I know that cos(φ) is equal to adjacent over hypotenuse.
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That's the adjacent side to φ, so φ is over here, it's adjacent side is 3, hypotenuse is still 5.
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So, φ=arccos(3/5).
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Again, I'll do that on my calculator.
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The calculator tells me that that's 53.1 degrees, approximately equal to 53.1 degrees.
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Now, there's a little check here you can do to make sure that you work this out correctly.
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We know that the angles of a triangle add up to 180 degrees.
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If we check here, 36.9+53.1 plus the last angle was a 90-degree angle, a right angle, if you add those up, 36.9+53.1 is 90 plus 90, you get 180 degrees.
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That tells me that my answers are right.
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The key formula here to remember is SOH CAH TOA.
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Everything comes down to drawing the angles in the triangle, and just using sine equals opposite over hypotenuse, cosine equals adjacent over hypotenuse, tangent equals opposite over adjacent.
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Now, we're given a right triangle and we're told that one angle measures 40 degrees.
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Let me call that angle right here, that would be 40 degrees.
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The opposite side has length 6.
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I want to find the lengths of all the sides in the triangle.
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Of course, finding the angles is no big deal because one side is a right angle, we're told that it is a right triangle.
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I can find the other angle just by subtracting from 180.
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In fact, 180-40 is 140, minus 90, is 50.
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I know that other angle is 50.
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The challenge here is to find the lengths.
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We're going to use SOH CAH TOA.
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I'm going to apply SOH CAH TOA to 40.
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I know that sin(40) is equal to 6 over the hypotenuse, because that's opposite over hypotenuse.
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The hypotenuse, if I solve this, that's equal to 6/sin(40).
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Remember to convert your calculator to degree mode before you do this kind of calculation.
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6/sin(40) is approximately equal to 9.3.
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That tells us the length of the hypotenuse, 9.3.
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Now I'd like to find the length of the other sides.
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I'm going to use the cosine part of SOH CAH TOA.
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I know that cos(40) is equal to the adjacent over the hypotenuse.
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If I solve that for the adjacent side, that's equal to hypotenuse times the cos(40).
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I know the hypotenuse now.
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If I multiply that by cos(40).
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What I get is approximately 7.2 for my adjacent side.
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The sides of my triangle are 6, 7.2 and 9.3.
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Again, there's an easy way to check that.
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We'll check that using the Pythagorean theorem.
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I want to check that 6²+7.2² gives me 87.8, which is approximately equal to 9.3².
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I know that I got those side lengths right.
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The third example here, we have the lengths of the two short sides of a right triangle, are in a 5:2 ratio.
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Let me draw that out.
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We're given that it's a right triangle.
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We've got lengths in a 5:2 ratio.
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I don't actually know that these lengths are 5 and 2, but here's the thing, if I expand this triangle proportionately, it won't change the angles.
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If I blow this up to a similar triangle that actually has side lengths of 5 and 2, I'll get a similar triangle with the same angles.
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I can just assume that this triangle has actually side lengths of 5 and 2.
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I want to find all the angles in the triangle.
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I'll label that one as θ, that one is φ.
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I know that sooner or later, I'm going to need the hypotenuse of the triangle, I'll go ahead and find that now.
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h²=2²+5², that's equal to 4+25 which is 29.
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My hypotenuse is the square root of 29.
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I'd like to practice all parts of SOH CAH TOA, and we've used sine and cosine in the previous problems.
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I'm going to use the tangent part.
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Remember, tangent is opposite over adjacent.
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Tan(θ) there, the opposite side is 5, and the adjacent side has length 2.
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I'm going to find arctan(5/2), θ is arctan(5/2).
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Remember that you want to have your calculator in degree mode here, because if you have your calculator in radian mode, you'll get an answer in radian which would look very different from any answer in degrees that you were expecting.
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I calculate arctan(5/2), and it tells me that that is approximately equal to 68.2 degrees.
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Now, I found θ, I've got to find φ now.
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I could find φ just by subtracting θ from 90 degrees, because I know θ+φ adds to 90 degrees.
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Remember, one angle of the triangle is already 90 degrees, the other two must add up to 90 degrees.
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I could just find the other angle by subtracting but I want to avoid that.
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I want to practice using my SOH CAH TOA rules.
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The other reason is if I find it using some other method, then I can add them together at the end and use that to check my work.
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I'm going to try using a SOH CAH TOA rule.
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I'm going to find it using angle φ.
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Looking at φ, I know that sin(φ) is equal to the opposite over hypotenuse.
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Sin(φ) is equal to the opposite over hypotenuse.
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The opposite side to φ is 2, and the hypotenuse is the square root of 29.
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φ is equal to the arcsine of 2 over the square root of 29.
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That's definitely something I want to put into my calculator.
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I'll figure out inverse sine of 2 divided by square root of 29 ...
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The inverse sine of 2 divided by square root of 29 ...
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It tells me that that's approximately equal to 21.8 degrees.
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That's what I got using SOH CAH TOA.
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Again, I'm going to check it by checking that the three angles of this triangle add up to a 180 degrees.
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I've got 68.2 degrees plus 21.8 degrees, those add up to 90, plus the last 90-degree angle, the right angle.
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Those do add up to 180 degrees.
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That tells me that my work must probably be right.
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That came back to looking at SOH CAH TOA, and figuring out what the angles were based on the SOH CAH TOA.
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We know that tangent is opposite over adjacent, and we also used that sine is opposite over hypotenuse.
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We'll try some more examples later.