hey class Miss Anderson here today I'll walk you through some right triangle trig review so most of this will be review we will learn a couple new trig ratios today but let's first recall the ones you've already learned in Geometry we have S cosine and tangent and we have S is going to be your opposite side over your hypotenuse coine is the adjacent side over the hypotenuse and tangent is opposite over adjacent you can this and write those down we will have a diagram here later on and show you which sides go with which again to remind you remember we learned sooa to help us remember the definitions for S cosine and tangent so soaa stood for where s is opposite over hypotenuse cosine is adjacent over hypotenuse and tangent is opposite over adjacent so hopefully that looks familiar we also will use the Pagan theorem which says the sum of the S squared equals the hypotenuse squared so that's another useful one to remember and so the legs it doesn't matter which one is a and which one is B but the hypotenuse is our C and that's a squ plus b s = c^2 and the new ones you're going to learn today are the reciprocal functions for S cosine and tangent so cosecant is the reciprocal of s So Co secant is the same thing as 1 / s of an angle and since it's the reciprocal s is opposite over hypo hypotenuse so cosecant is going to be hypotenuse over opposite secant is the reciprocal of coine so secant goes with cosine so since cosine is adjacent over hypotenuse you'll find the secant value by taking hypotenuse over adjacent and then we have coent that's the easiest one to remember because it has the word tangent in it it is the reciprocal of tangent which is adjacent over opposite so those are the three new ones let's look at it with a triangle so we have our triangle here it's a right triangle and this is our angle our sides are labeled based off of where the angle is located so the side that is touching the angle is considered to be the adjacent side that's what the word adjacent means the side that is across the leg that is across from the angle is considered to be the opposite side and then the hypotenuse is always the largest side and it's always across from the largest angle which is your 90° angle so adjacent and opposite might change based on where your angle is located at and then the hypotenuse is always across from the 90° so if I draw the same triangle but I move my angle to up here then this becomes my opposite this becomes my adjacent and the hypotenuse is still the the across from the 90 so to go with the the six ratios then you'll find the sign of an angle by taking the opposite side over the hypotenuse so this is our basic so CA and sometimes when I write this I write it like this so TOA so that reminds me that s is opposite over hypotenuse cosine is adjacent over hypotenuse and tangent is opposite over adjacent and then the new ones are the reciprocal functions so cosecant which is abbreviated as CSC this one is called cosecant goes with s so it's the inverse of or the reciprocal of sign so it's hypotenuse over opposite cant which is abbreviated secc goes with cosine so it is the reciprocal hypotenuse over adjacent and then Co is the abbreviation for coent and it's the reciprocal of tangent so it's the adjacent side over the opposite side so take a second make sure you get those down correctly then we're going to go on and use this to evaluate the six trig functions given the sides of a right triangle so our angle is in the top right corner so the side across from it is going to be our opposite side the side next to it is going to be our adjacent side and then our longest side and also the side across from the right angle is going to be our hypotenuse so we find S is going to be opposite over hypotenuse adjacent over hypotenuse and then opposite over adjacent for those three main ones so s will be at 1213 because the opposite side is 12 the hypotenuse is 13 which means the cosecant CSC will be the reciprocal 13 over 12 cosine is adjacent over hypotenuse so that's going to be 5 over 13 and secant is the reciprocal of that so 13 over 5 we're going to leave our fractions as improper fractions so 13 over five not a mixed number we also would simplify if we could and we want to leave it in fraction form ratio of side lengths not as a decimal tangent is going to be opposite over a hyp opposite over adjacent so 12 over 5 and coent will be the reciprocal 5 over 12 again leaving them as ratios of side lengths no decimals no mixed numbers Etc okay we're going to do this again but this time I gave you the legs of the hypotenuse and or the legs of the right triangle and not the hypotenuse so you'll have to use the Pythagorean theorem to solve for the hypotenuse it doesn't matter this one could be a this one could be B this one could be a b this one could be a that doesn't matter what matters is that your hypotenuse is C that's the that's key and then the um Pythagorean theorem is a 2 + b^2 = c^2 so the sum of the square of the legs equals the hypotenuse squar so that would give me 10^ 2 + 5^ 2 = c^ 2 that's 100 + 25 = c^ 2 so the square root of 125 is going to give me five square roots of five in simplified radical form so 5qu roots of 5 now my angle is in the top right corner again so this 10 is going to be my opposite side the five is touching it so that's my adjacent side and then this is my longest side if you know you if you did do the decimal the square root of 125 you're going to see it's going to be 11.18 so it is bigger than both of the legs the hypotenuse is always the biggest side so soaa sign is going to be o over H C is a over H and tangent is opposite over adjacent so sooa so s will be opposite over hypotenuse now we're going to reduce this a little bit and rationalize it so first 10 / 5 simplifies to 2 over 1 and then we got to multiply by theare < TK of 5 on both the top and bottom to rationalize our answer so we're going to get 2 square roots of 5 over 5 as a final answer there now if we go for our reciprocal function go back to the original and take the reciprocal of that so 5 square roots of 5 over 10 then simplify 5 over 10 is 1/ 2 so sare < TK of 5 over 2 for that one cosine is adjacent over hypotenuse so 5 over 5 roots of 5 again we can simplify 5 / five simplifies to one and then rationalize and then we get 5 squ < TK 5 * 1 5 over 5 and then for secant take the reciprocal of the original so hypotenuse over adjacent and then simplify to square < TK of 5 and then tangent is opposite over adjacent and 10 / 5 simplifies to two coent would be adjacent over opposite and that's going to simplify to 1 over two or2 so you will have to practice these and make sure you know which ones go with which I've lined these up nicely but they are not lined up nicely in your homework or in your workbook so make sure you pay attention to which ones go with which so that you get used to identifying the reciprocal functions correctly we can also use right triangle trig to help us solve a right triangle a solving right triangle means finding all of the missing side lengths and angles so we have one side length and one angle finding the missing angle is the easiest because we know that the angles in a right triangle add up to 180° and we have a 90° angle there so that means that angle a and angle B must equal 90 together so if you take 90 minus 62 you will get 28 and that's going to be your measure of angle a so that's one of our answers okay and we know the measure of angle B and then we can solve for our B and C values when I'm solving a triangle I always stick with the angle that was given and I label my triangle based off of that so 62 degrees was given which means B would be my opposite six would be my adjacent and C would be my hypotenuse and hi I got that email from you I think I lost that sheet I'll take it again I'll see them on Monday thanks okay sorry about that so we are going to use that as the side lengths and then I'm going to set up a trig um functions to help me solve for the opposite and the adjacent so if I want to solve for one of these if you think about so TOA I am given and currently the adjacent side whoops so I can either use cosine to solve for the hypotenuse or tangent to solve for the opposite so I'm going to start with Tangent I can't use sign of that angle because I don't know the opposite and I don't know the hypotenuse so you got to pick one that you know so if I do tangent of of the angle 62° would be equal to B over 6 got to try to make my B's and My Sixes look different then solve this for B this is divided by six so I would multiply by six on both sides to cancel that and that gives me what b equals so then make sure that your calculator is in the right mode you want to be in degree mode so if radians is highlighted you have to switch it hit enter and then hit exit you got to do 6 * the tangent of 62 you should get 11.3 about okay so there's another answer and then we have to solve for C you could use the Pythagorean theorem here you would want to write down a couple more decimals so you don't round too early if you're going to use the Pythagorean theorem or we can use cosine so cosine of angle 62 would be equal to adjacent side which is 6 over the hypotenuse which is C there's two steps involved in solving this one though because first you have to multiply by C to get C out of the denominator over here and then we get C * the cosine of 62 = 6 so then to get C by itself we have to then divide by the cosine of 62 on both sides so it ends up being C is 6 / cosine of 62° which is approximately 12.8 and now I found my missing side lengths and my missing angle and so the triangle is solved all right we're going to do it again but this time we're only given two side lengths and so we need to find the two angles and we need to find the missing side length the missing side length happens to be our hypotenuse so we can label these A and B in any order and use the Pythagorean theorem to find our missing side length so 11 2 + 18 2 = c^ 2 121 + 324 which is 445 the square OT of 445 is approximately 21.1 so that's what we'll say the side length is and then to find the missing angles we have to use inverse trig and so if we looking for angle P here angle P 18 would be opposite and this would be adjacent to angle P so if I wanted to find the measure of angle P I would say that the tangent of angle p is equal to 18 over 11 and then since what we know about inverse functions is that they switch the input and output switches when you do the inverse and remember your inverse notation was to the -1 so if you find that button that's the inverse then you switch your input and your output so when you type in the inverse tangent of 18 over 11 it will actually give you the angle measure the inverse buttons on this calculator are just above the regular buttons and notice the inverse notation is to the 1 power so that's what you're looking for so 18 over 11 gives me p is about 58.6 de and then we could do the same thing for angle n but if we look at angle n then this is my opposite side and this is my adjacent side so tangent of angle n would be 11 over 18 so if we did the inverse we could switch the input and output and that would give us our angle measure or I guess you could have took the um angle that you found for p and subtracted it from 100 or from 90° because these two angles should make up 90° together so angle n was 31.4 and 58.6 which does add to 90 CU we want a total of 180° in that triangle all right all right one more right triangle trig is often used in application problems as well so for example if you measure a flag pull the classic flag pull problem so you got a flag here whoops and you are standing 20 ft from the base of the flag we're assuming the flag is making a 90° angle with the ground angle of elevation is the angle that's formed from the horizontal up so that's right here 65 find the height of the pole so angle of elevation is 65 from where you're standing 20 ft away find the height of the flag pole So based off of the 65° angle that was given I have looking for the opposite side given the adjacent side opposite over adjacent goes with Tangent so that's what you need to use tangent of 65 will be equal to X over 20 and then solve for x by since it's divided by 20 the inverse operation would be to multiply by 20 so 20 * the tangent of 65 is going to be [Music] 42.9 ft okay hopefully that didn't get too long lots of things to cover there have a great day thanks for watching