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## High school geometry

### Course: High school geometry>Unit 5

Lesson 8: Sine & cosine of complementary angles

# Trig challenge problem: trig values & side ratios

Sal is given a diagram with multiple right triangles and is asked to match different expressions with different trig values. Created by Sal Khan.

## Want to join the conversation?

• I'm confused. How is cos(<DEC) equal to sin(41°)?
(10 votes)
• Cos(<DEC)=Sin(41˚) because cos (<DEC)=adj/hyp which is also = sin(41˚) = opp/hyp
(12 votes)
• does sohcahtoa only apply to right triangles?
(4 votes)
• sohcahtoa can be only applied in right angled triangles.
(15 votes)
• Around , how do you know the angle he marked was the one you were supposed to take the cosine of? Because the of the order of the letters?
(6 votes)
• Yes the order of the letters mean that the vertex of the angle will be the central letter, so you can trace the letters in order to create an angle.
(6 votes)
• Where could trig be used in the real world?
(5 votes)
• It's used in a plethora of occupations such as architecture, engineering, computer programming, graphical design, manufacturing, construction, and much more. Trig serves as a huge part of the foundations of Geometry, so basically anything in life that involves shapes finds uses for Trigonometry.
(7 votes)
• Am I correct in assuming that if we have a right triangle where the other two angles are x and y then we will always get cos(x) = sin(y) and cos(y) = sin(x)?
(5 votes)
• Yes! Because cos is A/H,sin is O/H and if θ is an angle and θ+90 is another,in a right triangle ,then the side opposite of θ will be adjacent to θ+90,and the side opposite of θ+90 will be adjacent to θ.
(5 votes)
• Am I capable of determining the measure of each angle in the video without having to do so much work?
(4 votes)
• Hi Jorge,

Take heart. With practice you will be able to solve these problems very quickly. If it helps consider that today's work is the foundation for the future. Many interesting fields of study depend on trigonometry.

Regards,

APD
(5 votes)
• Forgive me if this was covered, but I don't remember it actually getting mentioned. So to clarify, when Sal takes the cos(∠DEC), I notice that he used the value of angle E. Is there a notational standard that dictates, that in a case like the one above, the reference angle will be the middle angle. I notice he also did this in the problem sin(∠CDA), where he used D as his reference. Thanks in advance for clarifying this. :)
(3 votes)
• Yes, the middle letter in that notation is the referenced angle by convention :).
<ABC and <CBA both reference the same angle.
(5 votes)
• Where do we use sin cos tan in real life
(2 votes)
• every time you use your GPS system to get somewhere, builders and surveyors, sailing, etc.
(3 votes)
• When Sal writes EC/DE can that be the same as EC/ED ? @ ish
(1 vote)
• Yes. It doesn't matter which letter you put first. So EC = CE and ED = DE.
(3 votes)
• What do you call triangles that don't have right angles? But you can use Sine Law and Cos Law.
(1 vote)
• Oblique triangles are non-right triangles.
(3 votes)

## Video transcript

Sort the expressions according to their values. You can put any number of cards in a category or leave a category empty. And so we have this diagram right over here, then we have these cards that have these expressions. And we're supposed to sort these into different buckets. So we're trying to say, well what is the length of segment AC over the length of segment BC equal to? Which of these expressions is it equal to? And then we should drag it into the appropriate buckets. So to figure these out, I've actually already redrawn this problem on my little I guess you call it scratch pad or blackboard, whatever you want to call it. This right over here is that same diagram blown up a little bit. Here are the expressions that we need to drag into things. And here are the buckets that we need to see which of these expressions are equal to which of these expressions. So let's first look at this. The length of segment AC over the length of segment BC. So let's think about what AC is. The length of segment AC. AC is this right here. So it's this length right over here in purple over the length of segment BC, over this length right over here. So it's the ratio of the lengths of two sides of a right triangle. This is clearly a right triangle, triangle ABC. And I could color that in just so you know what triangle I'm talking about. Triangle ABC is this entire triangle that we could focus on. So you could imagine that it's reasonable that the ratio of two sides of a right triangle are going to be the sine of one of its angles. And they give us one of the angles right over here. They give us this angle right over here. You say, well all they did is mark that angle. But notice, one arc is here, one arc is here. So anywhere we see only one arc, that's going to be 30 degrees. So this is 30 degrees as well. You have two arcs here, that's 41 degrees. Two arcs here, this is going to be congruent to that. This over here is going to be 41 degrees. This is three arcs. They don't tell us how many degrees that is, but this angle with the three arcs is congruent to this angle with the three arcs right over there. So anyway, this yellow triangle, triangle ABC, we know the measure of this angle is 30 degrees. And then they give us these two sides. So how do these sides relate to this 30 degree angle? Well, side AC is adjacent to it. It's literally, it's one of the sides of the angle that is not the hypotenuse, so let me write that down. This is adjacent. And what is BC? Well, BC is the hypotenuse of this right triangle. It's the side opposite the 90 degrees. So this is the hypotenuse. So what trig function, when applied to 30 degrees, is equal to the adjacent side over the hypotenuse? Let's write down sohcahtoa just to remind ourselves. So soh cah toa. Sine of an angle is opposite over hypotenuse, cosine of an angle is adjacent over hypotenuse. So cosine, let's write this down, cosine of 30 degrees is going to be equal to the length of the adjacent side. So that is AC, over the length of the hypotenuse, which is equal to BC. So this right over here is the same thing as the cosine of 30 degrees. So let's drag it in there. This is equal to the cosine of 30 degrees. Now let's look at the next one. Cosine of angle DEC. Where is DEC? So DEC-- D, E, C. So that's this angle right over here. I'll put four arcs here so we don't get it confused. So this is angle DEC. So what is the cosine of DEC? Well once again, cosine is adjacent over hypotenuse. So cosine of angle DEC, the adjacent side to this, well that's this right over here. You might say, well isn't this side adjacent? Well that side, side DE, that is the actual hypotenuse. So that's not going to be the adjacent side. So the adjacent side is, I could call it EC. It's the length of segment EC. And then the hypotenuse is this right over here. The length of the hypotenuse-- the hypotenuse is side DE, or ED, however you want to call it. And so the length of it is, we could just write it as DE. Now what is this also equal to? We don't see this choice over here. We don't have the ratio EC over DE as one of these choices here. But what we do have is one of-- we do get one of the angles here. They give us this 41 degrees. And the ratio of this green side over-- the length of this green side over this orange side, what would that be in terms of, if we wanted to apply a trig function to this angle? Well, relative to this angle, the green side is the opposite side, and the orange side is still the hypotenuse. So relative to 41 degrees-- so let's write this down-- relative to 41 degrees, this ratio is the opposite over the hypotenuse. It's the cosine of this angle, but it's the sine of this angle right over here. Sine is opposite over hypotenuse. So this is equal to the sine of this angle right over here. It's equal to the sine of 41 degrees. So that is this one right over here. The sine of 41 degrees. So let's drag that in to the appropriate bucket. Sine of 41 degrees is the same thing as the cosine of angle DEC. Only have two left. So now we have to figure out what the sine of angle CDA is. So let's see, where is CDA? CDA is this entire angle. It's this entire angle right over here. So I could put a bunch of arcs here if I want just to show that it's different than all the other ones. So that's that angle right over there. So now we're really dealing with this larger right triangle. Let me highlight it in this pink color. So we're now dealing with this larger right triangle right over here. We care about the sine of this whole thing. Remember, sine is opposite over hypotenuse. So the opposite side is going to be side CA. So this is going to be equal to the length of CA over the hypotenuse, which is AD. So that is going to be over AD. And once again, we don't see that as a choice here. But maybe we can express this ratio-- maybe this ratio is a trig function applied to one of the other angles. And they give us one of the angles. They give us this angle right over here. I guess we could call this angle DAC. This is 30 degrees. So relative to this angle, what two sides are we taking the ratio of? We're taking, now, the ratio of, relative to this angle, the adjacent side over the hypotenuse. So this is the adjacent side over the hypotenuse. What deals with adjacent over hypotenuse? Well, cosine. So this is equal to the cosine of this angle. So this is equal to cosine of 30 degrees. Sine of CDA is equal to the cosine of this angle right over here. So this one is equal to this right over here. So let me drag that in. So this one is equal to-- so you can see that I just dragged it in-- equal to that. Now we have one left. We have one left. Home stretch, we should be getting excited. AE over EB. AE, let me use this color. Length of segment AE. That's this length right over here. Let me make that stand out more. Let me do it in this red. This color right over here, that's length of segment AE, over length of segment EB. This is EB right over here. This is EB. So now we are focused on this right triangle right over here. Well, we know the measure of this angle over here. We have double arcs right over here, and they say this is 41 degrees. So we have double marks over here, and this is also going to be 41 degrees. So relative to this angle, what ratio is this? This is the opposite over the hypotenuse. Opposite over the hypotenuse, this right over here is going to be sine of that angle, sine of 41 degrees. So it's equal to this first one right over there. So let's drag it. So this is going to be equal to sine of 41 degrees. So none of the ones actually ended up being equal to the tangent of 41 degrees. Now let's see if we actually got this right. I hope I did. We did.