Suppose ABC is a right triangle with sides of lengths a, b, an...

Question

Suppose ABC is a right triangle with sides of lengths a, b, and c and right angle at C. Use the Pythagorean theorem to find the unknown side length. Then find exact values of the six trigonometric functions for angle B. Rationalize denominators when applicable. See Example 1. a = √2, c = 2

Accepted Answer

Hey, everyone in this problem, we're asked to determine the missing side in the given triangle using the Pythagorean theorem. And also to write the value of all six trigonometric ratios for angle P. So we're given a triangle P. QR we're told that side P is equal to the square root of seven and side Q is equal to seven and Q is our right angle. And our hypotenuse when we're talking about this. So using the Pythagorean theorem, OK, we have the, the hypos which is Q squared is gonna be equal to the sum of the squares of the other two sides. So Q squared is equal to R squared plus P squared. We're told that Q is seven. So we have seven squared is equal to R squared plus P square, which is the square root of seven squared. Simplifying we have 49 and equal to R squared plus seven. This tells us that R squared is equal to 42. OK? By subtracting both sides by seven. And finally, if we take this square root, we get that R is equal to the square root of 42. And we're taking the positive route because we're talking about a distance. So we only care about those positive values. All right. So we found the missing side. That was the first step. So that's done. And let's move on to part two of this question. In part two is to write the value of the six Trigg ratios for angle P. All right. So let's just draw on our, on our diagram here just to indicate where angle P is. So we don't get confused and let's start with the sign. OK. And recall that sign, the angle P is gonna be equal to the opposite side divided by the hips. OK. So the opposite side to angle P is going to be side P and the partners we know is side Q. So we have P divided by Q which is gonna be equal to the square root of seven divided by seven. All right. So we have sign of P. Let's move now to, yeah, we're just gonna go through all six of them and cosine of P is going to be equal to the adjacent side divided by the hypo. OK. Let's look at our diagram. The adjacent side to angle P is going to be R. The hippo is still gonna be cute. So this is gonna be R divided by Q. So we found that R is equal to the square root of 42. So cosine of P is gonna be the squared of 42 divided by six all right, two down or to go next is, and we can do tangent a couple of different ways. OK? We can do tangent by knowing that it's the opposite side divided by the adjacent side. We can also do this by saying tangent of P is gonna be sign of P divided by cosine of P. OK. Either way works, either way is gonna give you the same answer. Now, in this case, sin of P we found was the squared of seven divided by seven cosine of P is the square root of 42 divided by seven. So what we get is the square root of seven divided by seven, multiplied by seven divided by the square root of 42. Those sevens will divide them. We're left with the square root of seven divided by the square root of 42. And we can simplify, OK. 42 and seven are both divisible by seven. We can divide both by seven and we end up with one divided by the square root of six. We want to rationalize that denominator. We're gonna multiply the numerator and denominator by the square root of six. And so we end up with the squared of six divided by six. Yeah. All right. So that one was a little bit more simplification that needed to happen, but it's the exact same process. OK. Tangent of P is squared to six divided by six. We're halfway there. Now we're on to CO C can OK. And CO C can FP we know is related to sine. So we have co C can P is equal to one divided by sine FP. This is gonna be one divided by route seven, divided by seven. This gives us seven divided by the square root of seven. And just like what we did with tangent of P, we want to rationalize the denominator. We don't want that square root in the denominator. So we're gonna multiply the numerator and denominator by the square root of seven that gives us seven multiplied by the square of seven divided by seven. And we can simplify that just to the square root of seven. That's now moving to care CK is going to be equal to cos or, or one divided by cos cosine of P. Sorry. OK. So if CK equals one divided by cosine of P again, we found cos P, let's go ahead and use that. We have one divided by the square root of 42 divided by seven. This gives us seven divided by the square of 42. Once again, we want to rationalize that denominator. We're gonna multiply it numerator and denominator by the squared of 42. So we have seven route 42 divided by 42. And we can simplify by dividing numerator into nominator by seven. And we get the square root of 42 divided by six. So that's CK of P squared of 42 divided by six. And we're on to the final trig ratio. We have cotangent of P which is going to be equal to one divided by tangent of peat. Right now. For tangent of P, we got the square root of six divided by six. We're not gonna use that. We're actually gonna use one step before that to make things a little bit easier. One step before that we had one divided by the square root of six. We're gonna use that value. OK? And we're gonna use that value so that when we take the reciprocal like we need to for code tangent, we're gonna end up with just the square root of six and we won't have to do any further simplifying. That's gonna be it. All right. So let's recap here. We found our missing sideline to be the square root of 42. We found that sine P is equal to the squared of seven divided by seven cosine of P is the squared of 42 divided by seven tangent of P is the square root of six divided by six co C can of P is the square root of seven D can of P square root of 42 divided by six. And finally, the code tangent of P is equal to the square root of six. Thanks everyone for watching. I hope this video helped see you in the next one.

Suppose ABC is a right triangle with sides of lengths a, b, and c and right angle at C. Use the Pythagorean theorem to find the unknown side length. Then find exact values of the six trigonometric functions for angle B. Rationalize denominators when applicable. See Example 1. a = √2, c = 2

Key Concepts

Pythagorean Theorem

Trigonometric Functions in Right Triangles

Rationalizing Denominators

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