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# Trig challenge problem: arithmetic progression

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if the angles A, B, and C of a triangle are in an arithmetic progression and if a, b and c - lower case a, b and c denote the length of the sides opposite to the capital, the angles A - capital A, capital B and capital C respectively then what is the value of this expression right over here so let's if we can work our way through this, so let's just draw the triangle so we have visualization of what all the letters represent so we have the angles A, B and C so let me just draw like this so we have the angles A B and C and then the sides opposites them are the lower case versions, so the side opposite the capital A is lower case a the side opposite to capital B is lower case b and the side opposite to the angle capital C is lower case c. Now the first piece of information they tell us is that they the angles capital A, capital B and capital C of the triangle are in an arithmetic progression very fancy word but all an arithmetic progression is, is a series of numbers that are separated by the same amount, so and let me give you some examples, so 1,2,3 - that's an arithmetic progression 2, 4, 6 arithmetic progression were separated by 2 every time I could do 10, 20, 30 also an arithmetic progression. These are all arithmetic progressions. So all they're saying is is to go from angle A to angle B - however much that is, is this the same amount to go from angle B to angle C. So let's see what that tells us what that tells us about or maybe tells us, maybe doesn't tell us anything about those angles so if we could say we could say we have angle A and then we have the notion angle B so we could say that B is equal to A plus some constant we don't know what that is. it could go by 1, it could go by 2, it could go by 10 we don't know what it is, so A plus N and then C would be equal to B plus N which is the same thing which is the same thing as B is A plus N, so this is A plus N it plus plus N which is equal to A plus 2N so what does that us? Well, the other thing we know about three angles in a triangles that they have to add up to a 180 degrees. So: this, this and this have to add up to a 180 degrees let's try it out so we have A plus A plus N plus A plus 2N, plus A plus 2N is going to be equal to 180 degrees we have one, two, three "A"s here, so we get 3A plus with one N and another two N: three A plus 3 N is equal to 180 degrees or you could divide both sides by 3 plus N is equal to and you get A A plus N is equal to 60 degrees so what does that what does that tell usŁż well A could still be anything cause if N is 1 then A is 59 if N is 10 then A is going to be 50, so it doesn't give us much information about the angle A but if you look up here, do you see A+N anywhere? you see it right over here. B is equal to A plus N and we just figured out that A plus N has to be equal to sixty degrees so using this first piece of information we are able to come up with something pretty tangible B must be equal to sixty degrees and you could try it out with a bunch of numbers that's an arithmetic progression and once again B is the middle one right over here. but no matter what the arithmetic progression is in order for these three angles to add up to 180 the middle one has to be equal to has to be equal to 60 degrees. So that was a pretty that's.. we're doing pretty well so far so let's see what we can do it the next part when the next part of problem. I'm trying to save some screen real estate right over here okay so they want us to figure out the value of the expression a over c sine of two C - capital C, plus c over a sine of two A. So let me just write it down so we have i'll do it.. i'll do it in blue a over c a over c sine of two times capital C plus c over a sine of two times capital A what's that going to be equal to? so, whenever you see stuff like this you got a 2 here a 2 here frankly the best things you do is just experiment with your trigonometric identities and see if anything pops out of you that might be useful and a little bit of a clue here the first part of the problem helped us figure out what B is it helped us figure out what B is but right now the expression has no B in it. So right now this information seems kind of useless but if we could put this somehow in terms of B then we'll have will will be making progress as we know information about angle B so let's see what we can do so the first thing i would use is with sine of 2A - let me just rewrite each of these so sine of i just say sine of two times anything that's just the same thing as and this is called the double angle formula so this is I might be wrong I was forgot the actual names of them but sine of two time something is two sine of that something times the cosine of that times the cosine of that something and you'll see that in any trigonometric book on the inside cover even a lot of calculus books let's do that for this the same thing right over here so sine of 2A over here is going to be 2 sine of A cosine of A that's just a standard trigonometric identity and we've - in the trigonometric that we prove that identity I think we do it multiple times than on front we have our coefficients still we have a over c times this plus c over a times this now is there anything we can do and remembered in the back of our mind we should be thinking of how can we use this information that B is equal sixty so if we can somehow put this in the form get B here when I think about how do you get a B here I think well you know we have a triangle here so the things that relate the sides of the triangle when especially it's not a right triangle we're gonna deal with the law of sines or law of cosines and law of sines let me just rewrite it over here just for our reference so law sines would say sine of A over a is equal to sine of B over b which is equal to sine of C over c at looks like we might be able to use that let me just write the law of cosines here just in case it's useful in the future so law of cosines c squared it's really the Pythagoras Theorem with little adjustment for the fact that it's not a right triangle so c squared is equal to a squared plus b squared minus 2ab cosine cosine of c - of capital C so it's law of sines, law of cosines. if we can somehow use both of these to put these in terms of b which we have information about well the first thing is I could rewrite this. So this is sine of c over c and this is sine of a over a So let me do that so I have let me do this. so I have the two a I have two a cosine of c. let me write that separately so I have two a cosine cosine of capital C and then times sine of C over c times I'll do it white. \4c&H000000&sine of C that's a capital C sine of capital C over lower case c that's that term and that term right over there and then to that I'm adding to that I'm adding - I'm doing the same thing over here I have two times I'm going to separate these guys out actually now I want to do the sines. So let me separate this guy and this guy out and so I get plus two c cosine of A times sine of capital over lower case a times sine of capital A over lower case a and I want this do for me Look that the sine right over there I have sine over c that's that over there and then I have sine of A over a. That's that over there, capital A over lower case a they're both equal to sine of B over b so we were making progress. We have - we started introducing B into the equation and that's the expression that we actually have information about so this to be rewritten as sine of B over b so this is the same thing as sine of capital B over lower case b and this is the same thing sine of capital B over lower case b and they are both being multiplied, or both of these terms are multiplying - be multiplied by that two a cosine of capital C times that and then plus two c it's a lower case c cosine of capital A times that so we can factor out the sine of B over b let's do that let's factor it out so this is the same thing as it's the same thing as two a and I really have a sense of what the next step is, so I leave a little space here two a times cosine of C plus, this - and these are being multiplied, I left some space there plus two c, two lower case c times a cosine of A and all of this all of this times the sine of B over b and we know - we already know the B is sixty degree so we can evaluate this uh... pretty pretty easily But let's just continue see if we can somehow somehow put this right over here in terms of b well if you look over here we have two a cosine of C, two c cosine of A it looks - it's starting to look pretty darn close pretty darn close, each of these terms are pretty darn close to this part to this part of the law of cosines over there and actually let's solve for that part of the law of cosines That's what we can do. so if you add two ab cosine of C on both sides, you get two ab cosine of capital C plus c squared is equal to a squared plus b squared or if you subtract c squared from both sides you get two ab cosine of capital C is equal to a squared plus b squared minus c squared and this is interesting and we can, you know - switch around the letters later on but this looks pretty darn close to this so what if - and this looks pretty darn close of this except we're here we're dealing with an a instead of a c. We just switch the letters around and we could rewrite this actually let me rewrite it just for fun I could rewrite this over here as two two c b not rewrite it. I can swap the letters times the cosine of A Here I'm swaping the as and cs, is equal to c squared plus b squared minus a squared there's nothing unique about sides c I can do this with all of the sides so here it's a big C here you have an a and a b out front and then you have the a square plus b squared minus the small c squared. if you have a big A, then you have the cb in the front and then you are subtracting the a squared right over here and this is useful because this term right over here this term right over here looks almost like this term over here if we could just - if we could just multiply this by b So let's do that we can multiply that by b but let's multiply this whole numerator this whole term by b so if we multiply this whole term by b. What do we get? we get a b there. we get a b there. and of course you can just arbitrarily multiply an expression by b that'll change its value so what we could do is multiply the expression by b which we just did we distributed a b across here but they'll also divide by b where which - so I'll divide by b. That's equivalent of multiplying the denominator the denominator there, not b squared that's equivalent of multiplying the denominator by b that's the same thing as dividing by b we multiplied by b, divided by b where that's the same thing as just turning this to b squared now what does this give us well we have this term right over here this term right over here is now the exact same thing as that over there so it is now a square plus b squared minus c squared and then this term right over here is now the exact same thing as this thing over here which is the same thing as that we are using the law of cosines so this is plus c squared plus b squared minus a squared and then all of that times this sine of B sine of capital B over b squared now what does this give us? we have an a square and a negative a square things are starting to simplify a squared, negative a squared we have a negative c squared and a positive c squared so what are we left is - so we just left the two b squared so our whole expression has simplified to two b squared sine of B sine of capital B over lower case b squared these cancel out so our whole expression simplifies to two sine of B and from the get go we knew what B was we know it's sixty degrees so this is equal to two times the sine of sixty degrees and if you don't have the sine of sixty degrees memorized you can always just break out a thirty sixty ninety triangle so let me draw - this is a right triangle right over here This is sixty degrees hypotenuse has length one we're dealing with the unit circle this side is thirty degrees the side opposite the thirty degrees is one-half the side opposite the sixty degrees is square root of three times that. So it's square root of three over two. you can even use the Pythagoras Theorem to figure out Once you know one of them you could figure out the other one the sine of - sine is opposite over hypotenuse so square root of three over two over one or just square root of three over two. It's equal to two times - it's the home stretch, we are exciting square root of three over two these cancel out. so we are left with the square root of three that's a pretty neat problem. and just in case you are curious where it came from, the two thousand and ten IIT, IIT are these - hard to get into uh... engineering and science universities in india and gives you the example like you know hundreds of thousands of kids and you know the top the top I don't know like two thousand actually get into one of IITs that I just thought it was a pretty neat problem