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## Class 10 (Old)

### Course: Class 10 (Old) > Unit 8

Lesson 2: Reciprocal trigonometric ratios# 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.

## Want to join the conversation?

- Does Soh Cah Toa only work for right angle triangles?(4 votes)
- Sometimes in a triangle-based problem that doesn't have a right triangle, you can cut the triangle into a couple of pieces forming right triangles, and then you can figure out the dimensions and/or angles using soh cah toa and logic to get to your victory lap. This is especially true if you are given the altitude of a triangle and the length of one of the sides connecting the altitude to the base. With triangles, there are often several ways to get to the answer.(25 votes)

- I am trying to understand from trig 1.5 questions how these legs give missing information in the form of an equation. To me it seems that when it asks for some of these numbers they are already used in the trig ratios but they are not the answer, it is like I have to compare two forms of the ratio, like it will ask for the adj, and there are three equations with information you need to use, which video shows you how to use this properly?(11 votes)
- IE

BC = 48

AC = ?

AC is the opposite of B, Angle ABC

Sin(angleABC) = 7/25

cos(angleABC) = 24/25

tan(angleABC) = 7/24

7 is not the answer, although SOH CAH TOA says it is a ratio of the opposite side over the hypotenuse. Why is 7 not the answer? I tried the same on some other problems and they were correct, this is not straight forward in some way. Why? What operations or way do I need to approach this problem to get missing information? The hints make it look like an equality is needed, but why when it seems apparent sin would give a relevant side of the ratio of the same triangle, or can it be expressed as some other information like on another similar triangle?(3 votes)

- How do you use sin, cos, tan, csc, sec, and cot to find angles?(7 votes)
- Well, finding angles is another major use of trigonometry. For example, if you know that the ratio between two sides is 1 to 2, and the sides happen to be opposite and hypotenuse to an unknown angle A, then you can use the sine ratio to tell you that sin A = 1/2. When an angle is equal to 30 degrees, then the sine ratio is always 1/2. So then you use known ratios to tell you that the angle has to be 30 degrees.

Likewise, the tangent (tan) of 45 degrees is 1, which means the opposite side over the adjacent side is 1/1 which is equal to 1. So if you have a tangent ratio equal to 1, you know the angle is 45 degrees. There are many common angles that you will become familiar with if you continue your study of trigonometry. These common angles are simple ones like 0 degrees, 45 degrees, 30 degrees, 60 degrees, 90 degrees, and simple multiples and fractions of these. In more advanced math, the angles are given in radians, but these basic angles are used so often that you will know them by heart.

Beyond those simple ones, you can use tables of the trig ratios or calculators to translate complicated ratios such as sin B = 0.2113 to find out that angle B is about 12.2 degrees.(7 votes)

- What is the difference between cosine and arcsine then? I mean it looks the same to me...(2 votes)
- The cosine and the arcsine are not the same at all.

The arcsine is the inverse function of sine. The cosine is the sine of the complementary angle.

With angles measured in radians, here are what each of these functions is equal to:

arcsine x = -i ln(ix+ √(1-x²))

cos x = ½ [e^(ix)+e^(-ix)]

Thus, there is no particular connection between the arcsine and the cosine.(13 votes)

- This question peaked in my head at5:31when the solution of the 6 trigonometric ratios are there. Is it possible that you could find all six trigonometric ratios if you just have the information about two trigonometric ratios or was this just an example that somehow had this?(5 votes)
- Actually, aside from worrying about the sign, you can find all six ratios given just one of them. Like Sal said at the end of the problem, you could have found the third side using the Pythagorean Theorem instead of the second ratio, or you could have used the inverse trig functions to find the measure of the angle and then applied all six functions to that angle. The only thing about the second ratio that was important to know is that the cotangent was positive -- if it was negative then some of the other ratios would have had to been negative as well.(5 votes)

- We're given a right triangle with two sides (5, square root of 41). Couldn't we use COS-1 of angle FDE to find the angle then SIN or TAN of angle FDE to find the length of the missing side? I don't think we even need the other information given to find the length of 'a'.(4 votes)
- You are correct, only one of the two given ratios is actually needed. You could also just use the Pythagorean Theorem to find the last side and go from there.(4 votes)

- I thought the reciprocal of a trig function like Sine was that function to the negative 1 power.(3 votes)
- A reciprocal is the inverse of the trig ratio. Sin^-1 is the inverse of sin itself. Good question.(2 votes)

- As angle theta increases from 0 degree to 90 degree , cos theta decreases from 1 to

0. why & how ?(2 votes)- You can know this if you draw the cosine curve. Also there is a way to find that if you differentiate the curve piece wise. Also try to find the limit at 0 and pi/2 . Unless you draw and analyse the curve you will not able to find it.(1 vote)

- Can you do a reciprocal of sin, cos, or tan on a calculator? I have do a question that ask: sec 10 degree

how do I do that on a calculator?(2 votes)- Some calculators will have designated buttons for sec, csc, and cot but most won't, in my experience. Then, all you have to remember is that these functions are the reciprocals of the normal trig functions. Putting this mathematically, for example:

csc(x) = 1 / sin(x)

So anytime you're asked to find secant, find the cosine of that angle and divide 1 by your answer.(3 votes)

- what's the difference between csc and sin -1?(1 vote)
- They are completely different. Sin⁻¹ x is the inverse function NOT the reciprocal of sin x.

This is an unfortunate, but standardized, inconsistency in math notation. If the`⁻¹`

comes immediately after the name/symbol for the function then it means "inverse function" NOT reciprocal. If the`⁻¹`

comes after a constant, variable or other expression then it means the reciprocal of the constant, variable or other expression.

You can avoid the problem by using arcsin x instead of sin⁻¹ x. But you must learn that any time you see function⁻¹ (variable, constant, other expression) you have the inverse function, NOT the reciprocal.(5 votes)

## 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!