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# Trig challenge problem: arithmetic progression

## Video transcript

if the angles a B and C of a triangle are in an arithmetic progression and if a B and C lowercase a B and C denote the lengths 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 see if we can work our way through this so let's just draw the triangle just so we have a visualization of what all of the letters represent so we have the angles a B and C so let me just draw them like this so we have the angles a B and C and then the sides opposite them or the lowercase version so the side opposite to capital a is lowercase a side opposite to capital B is lowercase B and the side opposite to the angle capital C is lowercase 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 an arithmetic progression 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 we're 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 it's the same amount to go from angle B to angle C so let's see what 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 up by one it could go up 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 plus n which is equal to a plus 2n so what does that do us well the other thing we know about three angles and the triangles that they have to add up to 180 degrees so this this and this have to add up to 180 degrees so 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 three a plus we have one N and then another two n 3 a plus 3 n is equal to 180 degrees or you could divide both sides by 3 and you get a plus n is equal to a plus n is equal to 60 degrees so what is that what does that tell us well it could still be anything because if n is 1 then a is 59 if n is 10 then a is going to be 50 so it doesn't to give us much information about the angle a but if you look up here do you see an A plus at anywhere well 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 60 degrees so using this first piece of information we were able to come up with something pretty tangible B must be equal to 60 degrees and you could try it out with a bunch of numbers these could be 59 60 and 61 that's an arithmetic progression and once again B is the middle one right over here these could be 50 60 and 70 could be 40 60 and 80 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 what we're doing pretty well so far so let's see what we can do it the next part with the next part of the problem and 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 2a so let's let me just write it down so we have I'll do it in I'll do it in blue a oversee a oversee 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 two here two here frankly the best things you do is just experiment with your trigonometric identities and see if anything pops out at 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 we'll we'll be making progress because we know information about angle B so let's see what we can do so the first thing I would use is well sine of 2a so let me just rewrite each of these so sine of I should say tyne of two times anything that's just the same thing as I think this is called the double angle formula so this is although I might be wrong there I always forget the actual names of them but sine of two times something is 2 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 or 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 in the trigonometric playlist we prove that identity I think we do it multiple times and then out in front we have our coefficient still we have a over C times this plus C over a times this now is there anything we can do and remember in the back of our mind we should be thinking of how can we use this information that B is equal to 60 so if we can somehow put this in the form get a B here and 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 a triangle when especially when it's not a right triangle is we're going to deal with the law of sines or the law of cosines and the law of sines let me just rewrite it over here just for our reference so the law of 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 and it 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 the law of cosines C squared it's really the Pythagorean theorem it's a little adjustment for the fact that it's not a right triangle so C squared is equal to a squared plus b squared minus 2a B cosine cosine of C of capital C so it's law of sines and law of cosines so let's see 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 says 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 2a I have 2a cosine of C let me write that separately so I have 2a cosine cosine of capital C and then times sine of C over C times I'll do that in white sine of C that's a capital C sine of capital C over lowercase 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 gonna do the same thing over here I have to times I'm going to separate these guys out actually now I want to do the signs so let me separate I'm going to separate this guy and this guy out and so I get plus 2 C cosine of a times sine of capital a over lowercase a times sine of capital a over a lowercase a now what did this do for me well look at the law so look at the law of sines right over there I have sine of C over C that's that over there and then I have sine of a over AI that's that over there capital a over lowercase a they're both equal to sine of B over B so we're making progress we have--we've relate we started introduced B into the equation and that's or the expression and that's what we actually have information about so this could be re-written as sine of B over B so this is the same thing as sine of capital B over lowercase B and this is the same thing as sine of capital B over lowercase B and they're both being multiplied or both of these terms are multiplying are being multiplied by that 2a cosine of capital C times that and then plus 2c2 lowercase e cosine of capital a times that so we can factor out the sine of b over b so let's do that let's factor it out so this is the same thing as this is the same thing as 2a 2a and I already have a sense of what the next step is I'm leave a little space here to a times cosine of C plus this and these are being multiplied I left some space there plus 2c2 lowercase e times the cosine times the cosine of a and all of this all of this times the sine the sine of B over B and we know we already know that B is 60 degrees so we can evaluate this pretty pretty easily but let's just 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 2a cosine of C - C cosine of a it looks it's starting to look pretty darn close pretty darn close each of these terms look pretty darn close to this part to this part of the law of cosines over there and actually let's let's solve for that part of the law of cosines let's see what we could do so if you add if you add to a B cosine C to both sides you get to a B 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 2a B 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 to this except here we're dealing with an a instead of a c we've just switched the letters around and we could rewrite this actually let me rewrite it just for fun I could rewrite this over here as 2 to see B not rewrite it I can swap the letters times the cosine of a so here I'm swapping the A's and the C's is equal to C squared plus B squared minus a squared there's nothing unique about the side see I can do this with all of the sides so here when you have a big C here you have an A and a B out front and then you have the a squared plus B squared minus the small C squared if you have a big A then you can have the C be in the front and then you're subtracting the a squared right over here and this is useful because this term right over here this term right over here it looks almost like this term over here if we just have to if we could just multiply this by B so let's do that we can multiply that by B right 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 to be there we get a B there and of course you can't just arbitrarily multiply an expression by B that'll change its value so what we can do is multiply the expression by B which we just did we distributed the B across here but then we'll also divide by B which so I'll divide by B that's the equivalent of multiplying the denominator the denominator there not B squared that's the equivalent of multiplying the denominator there by B that's the same thing as dividing by B we've multiplied by B divided by B or that's the same thing as just turning this into 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 squared 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're 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 the we have an a squared and a negative a squared things learning to simplify a squared negative a squared we have a negative C squared and a positive C squared so what are we left with we're just left with a 2b squared so our whole expression has simplified to 2b squared sine of B sine of capital B over lowercase B squared these cancel out so our whole expression simplifies to 2 sine of B and from from the get-go we knew what B was we know it's 60 degrees so this is equal to 2 times the sine of 60 degrees and if you don't have the sine of 60 degrees memorized you can always just break out a 30-60-90 triangle so let me draw this is a right triangle right over here this is 60 degrees hypotenuse has length 1 we're dealing with the unit circle the this side is 30 degrees the side opposite the 30 degrees is 1/2 the side opposite the 60 degrees is square root of 3 times at so it's square root of 3 over 2 you can even use the Pythagorean theorem to figure out once you know one of them you could figure out the other one so it's the sine of sine is opposite over hypotenuse so square root of 3 over 2 over 1 or it's just square root of 3 over 2 so this is equal to 2 times is the homestretch it's very exciting square root of 3 over 2 these cancel out so we are left with the square root of 3 that's a that's a pretty neat problem and just in case you're curious this came from the 2010 IIT IITs are these hard to get into engineering and science universities in India and they give you this exam to like you know hundreds of thousands of kids and you know the top the top I don't know like 2,000 actually get into one of the IITs but I just thought it was a pretty pretty neat problem