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

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.b = 8, c = 11

Accepted Answer

Everyone in this problem, we're asked to determine the missing side in the given triangle using Pythagorean theorem. And also to write the value of all six trigonometric ratios for angle P. So we're given triangle P. QR kq is the 90 degree angle. And so Q is also the hypo here, we're told that side R is equal to five and side Q is equal to 12. Now the missing side is side P, we have a right hand triangle. We're asked to use Pyar theorem. So let's get started with that the theorem or call tells us that the IPO square. So in this case, Q squared is going to be equal to the sum of the squares of the other two sides. So Q squared is equal to R squared plus P squared. Substituting in our values we have 12 squared is equal to five squared plus B squared that tells us that 144 is equal to 25 plus P squared. Subtracting both sides by 25 we get that P squared is equal to 119. And finally taking the square root P is going to be the square root of 109. And we take the positive route because we're talking about a distance here. All right. So part one of this is done, we have found the missing side. So check mark there. Now, I wanna write down these six trigonometric ratios. OK? And we're gonna start with side and recall that sign of the angle P is going to be equal to the opposite side divided by the hypo. Yeah, the opposite side from angle P. OK. Let's just I let angle P here. So we don't mix it up in our diagram. The opposite side is going to be side P. The hypotenuse is side Q. So sign of P is going to be equal to the square root of 119 divided by 12. All right, sp is that we move on to cosign. Now for cosine recall that for the trigonometric ratio cosine this is gonna be equal to the adjacent side divided by the hypo. OK. Looking at our triangle, the adjacent side is gonna be side R with a length of five. The hypotenuse is Q. And so we have that the cosine is gonna be R divided by Q which is equal to five divided by 12. All right. So we're done to four more to go. Yeah, but you see that right in these six tribe ratios, what we needed to do, we just needed to find those three side links. And now we just need to remember the tri ratios themselves. OK. Which sides they relate substitute those. Let's move to tangent. OK. In tangent of P, we can calculate in two different ways. OK. We can recall that the tangent is gonna be sine of P divided by cosine of P or we can also use that tangent of P is the opposite side divided by the adjacent side. OK? Either way you want to do it, the opposite side divided by the adjacent side. Well, that's gonna be P invited by art. And so we get that tangent is equal to the square root of divided by five. All right. Moving on now to cos seeking no CFP. We can do in a couple different ways just like tangent. OK? How we're gonna do it? Well, let's recall that Cohan of P is equal to one divided by sine of P. We don't need to do anything special. We don't need to remember anything special. We know what sign of P is. We just calculated it. Let's go ahead and use it. So is then going to be equal to one divided by the square root of 119. They're divided by 12. This is gonna be equal to 12 divided by the square root of 119. And we want to rationalize the denominator and we don't want the square root in the denominator. So we're gonna multiply both the numerator and denominator by the squared of 119. That's gonna give us 12 multiplied by the square root of 119 divided by 119. OK. And that can't be simplified any further. That's the expression for moving now to seek him. Now, C can of P, we're gonna do very similarly to how we did with CO C can, except this time it's related to cosine. So C can of P is gonna be one divided by cosine of P. This is gonna be one divided by 5 12, which is just equal to 12 5th. OK. So C can of P 12 divided by five. No more simplification that we can do there. And finally, we're on our last trigonometric ratio, we are going to calculate code tangent of P and just like with COCA and C camp, we are gonna use one of our other trig ratios, cotangent of P is equal to one divided by tangent of P. This is gonna be equal to one divided by the square root of 119 divided by five. This gives us five divided by the square of 119. And just like what we did earlier, we want to rationalize the denominator, we multiplied numerator and denominator by the squared of 119. And we get five multiplied by the squared of divided by 119. And that is the expression for co of P. So we found our missing side length. We found all six trig ratios. That's it for this problem. Thanks everyone for watching. I hope this video helped see you in the next one.

Struggling with Trigonometry?

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.b = 8, c = 11

Watch next

Trigonometric Functions on the Unit Circle practice set