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## Trigonometry

# Using reciprocal trig ratios

Sal is given two sides in a right triangle and the cotangent of one of the angles, and he uses this information to find the missing side. Created by Sal Khan and Monterey Institute for Technology and Education.

## Video transcript

For acute angle E, so that's this angle right over here, it tells that sin of E is 5 over the square root of 41, and cotangent of E is equal to 8/10. Find the values of the other four trigonometric ratios. So for the trig ratios, I like to use SOH-CAH-TOA to remember what the definitions of the trig ratios were So let me write this down, SOH, CAH I'll write CAH in a different color, SOH, CAH that's not a different color I'm trying- I'm having trouble changing colors! SOH, CAH, TOA. SOH, CAH, TOA. So, SOH tells us that sine of an angle in this case we care about angle E, is equal to the opposite over the hypotenuse. If we look at angle E, what is the opposite? Well, we go across the triangle the opposite is right over there. That is the opposite so it would be equal to 5 over the hypotenuse. Well, the hypotenuse is opposite the right-angle, and it's the longest side of the triangle, it's side DE and has length of square-root of 41. Square-root of 41. So that's consistently the information they gave us so this was kind of redundant information, we could have figured that out just from the diagram. But at least we got that out of the way. Now let's think about, now let's think about the reciprocal of the sine of E, which is the co-secant of E. Co-secant of E which is the reciprocal sine which is hypotenuse over the opposite. and we could- we don't even have to look at the triangle we could just say the reciprocal which would be the square-root of 41 over 5. Or you could look back at this and try to figure that out. Now let's think about the cosine. What is the cosine? What is the cosine of E going to be equal to? Well, first think of what the definition is. Cosine of E is going to be what? Well, CAH tells us that this is going to be equal to the adjacent over the hypotenuse. Well, if we look over here, we know what the hypotenuse is. The hypotenuse the same. The hypotenuse is the square-root of 41. What is the adjacent side? What length does the adjacent side have? Well, for angle E, the adjacent side is side FE and we don't know what that is right over here. So that is the adjacent side and we don't know what that is So I'll just write 'a'. 'a' for how they marked it on this, or you could also do this 'a' for 'adjacent. So right now we're just gonna leave it with the variable 'a' over the square-root of 41 maybe we can get a little bit more information that helps us to figure out what 'a' is over the course of this problem. If you wanna figure out what the secant of E is, well, that's just the inverse of the cosine it's hypotenuse over adjacent, so it's going to be the square-root of 41 over 'a'. Whatever 'a' is. Hopefully we can figure that out. Now let's use TOA. That tells us that the tangent of E is equal to the opposite over the adjacent. Well, what's the opposite side to angle E? Well, it's side- it has length 5, it's side DF The opposite is 5. And we still don't know the length of the adjacent side. That's this side of length 'a'. So I'll just write 'a', right over there. Now what about cotangent? Well, cotangent is the reciprocal of tangent. So it's adjacent over opposite or in this case, 'a' over 5. Adjacent over opposite; 'a' over 5. Well, what do they already tell us? And using that, can you figure out what 'a' is? Well, they already told us that cotangent of E is equal to 8/10. There's a little clue here that this is not reduced fully; 8/10. 8 and 10 share common factors. So they already tell us that cotangent of E is 8/10. So if we use the definition of cotangent of E we get 'a' over 5, and they tell us that that is going to be equal to 8/10. So we can write, we have an equation now to solve for 'a'. And if we solve for 'a' now we can figure out all of the other, all of the other ratios. So let's do that. So we have 'a' over 5 is equal to, -we'll simplify this in a little bit- what is 8/10 if you simplify it? Well, we have 8 by- we have a common factor of 2- if you divide 8 by 2 you get 4, we divide 10 by two, we get 5. So we get 'a' over 5 is equal to 4/5. And so this is actually- we could cross-multiply or multiply both sides by 5 and you would get -either way- you would get that 'a' is equal to 4. Let's do that, just to show you you can do it systematically. And you're left with 'a' is equal to 4 which is great because we can now say that the cotangent of E, yes it's 8/10 which is the same thing as 4/5. We could say that the tangent of E -instead of saying it's 5 over 'a'- we can now say that it's 5 over 4. And now what would the cosine of E be? Well, it was 'a' over (square-root of) 41, now it is -let me do that in a different color- now it is 4 over the square-root of 41. And what is the secant of E now? Well, it was square-root of 41 over 'a', now it's square-root of 41 over 4. Because now we know the value of 'a'. And we can verify that 'a' is equal to 4 by using the Pythagorean theorem. In fact, we could've solved it that way. But the whole point, I'm suspecting, of this problem, is to actually establish this (a = 4) using this (cotangent of E= 8/10) information even although we could have done it using the Pythagorean theorem. But let's just verify that this 'a=4' satisfies the Pythagorean theorem. So if we take this side right over here, we got 4 squared plus 5 squared should be equal to the square-root of 41 squared. Should be equal to the hypotenuse squared. Square-root of 41 squared. So 4 squared is 16, 5 squared is 25, so does this actually meet- does this actually satisfy the Pythagorean theorem? So 16 plus 25, square-root of 41 squared is 41, 16 plus 25 is indeed 41. So at least it's consistent with the Pythagorean theorem, and we are done!