 Welcome back. Today's topic is solving right triangles. The first thing we're going to do is just discuss what do we mean solving a triangle. Okay, that's kind of a new term. And it's actually just kind of a short phrase, just to say something a little quicker. And what it means is to determine the sides and the angles of a triangle. Usually we have to use some known information in order to do that. When we work with a right triangle, we always know the value of one of the angles, the measure of one of the angles, and we can then use, usually use the trigonometric ratios for an acute angle to help us find the other sides and possibly the angles of a right triangle. We can also use the Pythagorean theorem. And at some time when we might use the inverse trigonometric functions to help us with that. And so here's our usual setup for a triangle, and I tried to use this in very general notation. Instead of a specific one, when we have an angle, we generally talk about the side opposite the angle, the side adjacent the angle, and the hypotenuse, since we're working with a right triangle. And here's our definitions. If you want, within a triangle, we shorten these sometimes just to say opposite over hypotenuse for the sine of theta, adjacent over hypotenuse for cosine of theta, and adjacent, oops, sorry, opposite over adjacent for the tangent of theta. It would be possible to use cosecant and cotangent theoretically it's possible to use those. But in practice we use just these three when we're working with triangles, basically those are built in to our calculators and makes it a lot easier to do computations with those. And at times as I said we will also have to use the inverse trigonometric functions. So here's our first problem. We're given a certain angle, 52 degrees, 46 minutes, and the side adjacent to it is 20.3 meters. So one of the slight difficulties with this one is the angle being measured in degrees and minutes. And what I would do with that usually is just convert that to what I would refer to as decimal degrees. In my calculator I would do 52 plus 46 over 60 and hit enter. You will get something that's 52.76666 a bunch of sixes and ends with a seven. And that you don't want to have to type in all the time. So what I'm going to do is I'm going to store it, generally that little arrow there, in memory location theta. I use the arrow because basically if you hit store on your calculator a little arrow shows up on the calculator screen. This is talking about the TI-84 calculator. And theta is an alphabetic character built into the TI-84. You access that with the alpha key and then hit three. Look at that, you'll see theta above that. Now that's then stored in that and so any time I want that all I have to do is press theta. In other words, alpha three in order to get theta. And with that then I can start trying to solve this triangle. So we've got theta there and now I start introducing some symbols to help us with that. So we're going to let A be the length of the other side of the triangle. That one will be opposite of theta and I will let H be the hypotenuse. And now we can start looking at theta. And you can see this A is opposite and this is adjacent. That's a tangent. So in other words I can say A over 20.3 is tangent of theta. And solving that equation for A we get 20.3 times tangent of theta. And that gives us a chance now to use our calculator. We have to make sure, of course, our calculator is in degree mode because we're working with degrees and remember I'm using theta there. I could always type in 52 plus 46 over 60 instead but it's a lot easier just to type in A or theta and you'll find at least rounded to the nearest tenth this is 26.7. But before you do any other calculations take that answer of 26.7 and whatever the rest of that is and store that in memory location A. And you can see how I'm trying to set this up so I can easily recall where I have things stored. So if I call the unknown A I store it in memory location A. In the next slide we'll see a way to check our work and it's very nice to have the full calculator accuracy available to us to help us with the check. So now what I want to do is determine H. And you can see again I'll try to use this value always try to use the given information rather than a calculated value. Usually you get a little easier to work with the given information and that way you don't rely on a calculated value to determine something else. If you make a mistake with one calculated value it will continue for the other ones. So what I'm going to do again is look at this in relation to theta. This as we said before is the adjacent side and this is the hypotenuse. So if I do adjacent over hypotenuse I get cosine of theta. And now I have to solve for H. So the first step will be to multiply both sides of the equation by H. 20.3 equals H times cosine theta. And I'll write that as H will be 20.3 divided by cosine of theta. And again I now use my calculator and that can be done on the calculator. Again rounded to the nearest tenth which might be how I write down the answer. We get 33.6 but I'm going to store that the actual 10 digit number that I see on the calculator I'm going to store that in memory location H. So I've got A and H. The last thing is this angle up here which we will designate by phi and we can now use the fact that in any right triangle the sum of the two angles the two acute angles is 90 degrees. So phi plus theta is 90 or in other words phi is 90 minus theta. I can use the calculator to do that since I have theta stored. Let's just take a look at how we might do this with degrees and minutes. What I'm going to do is take the 90 degrees and write that as 89 degrees 60 minutes. Take one of the degrees and convert it to minutes. And then my given angle is 52 degrees 46 minutes. And now I can do the subtraction. And with subtracting the minutes we get 14 minutes. And the degrees we get 37 degrees. So there's the angle phi. Here are the answers we have obtained. And just to remember where we have stored things this has been stored in H. I'm sorry theta this has been stored in A and this has been stored in H. Not 33.6 but the full 10 digits that we obtained from the calculator for both H and A. And one way to check this work then is to use the Pythagorean theorem. And in particular if we take the A, the one side so I'll do A squared. Do that on my calculator from memory location A plus 20.3 squared. And I do that on my calculator. I get 1125.619798. I wrote out the whole thing there to kind of emphasize why this works well as a check. Now I'm going to go back and use memory location H for the hypotenuse and square that. And again that will give me 1125.619798. And after I've done that I'm very confident that I've got the correct answer for the side and the hypotenuse. Notice I could use the rounded values. This one of course was given to us. But if I use the rounded values I would do 26.7 squared plus 20.3 squared. And if you do that on the calculator it will come out to be 1124.98. And in hypotenuse squared again using the rounded value of 33.6 squared comes out to be 1128.96. That is not a perfect check. It probably indicates at least that I'm close to the correct answer and probably am correct. But this here is a much nicer check. It really makes it certain that we've got the correct answer there. So one more problem. And in this case we're given the lengths of two sides of the right triangle and we have to determine the hypotenuse and the two angles. So I'm going to introduce some notation. Again we'll call that H and I'll call the two angles alpha and beta. And again as much as possible we're going to try to use the given information to determine the other sides. So for the hypotenuse it's the Pythagorean theorem. So H squared will be 8.5 squared plus 12.2 squared. Doing that on a calculator.