More trig examples
IIT JEE Trigonometry Problem 1 2010 IIT JEE Paper I #29 Trigonometry problem
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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 see 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 them 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
- 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 tell us? Well, the other thing we know about three angles in a triangle
- 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 and you get A
- plus N is equal to
- A plus N is equal to 60 degrees
- so what does that
- what does that tell us
- 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 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.. we're doing pretty well so far so let's see what we can do at the next part
- with 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 should 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
- well 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 for 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, in the trigonometric, less we prove that identity i think we do it multiple times
- then out on front we have our coefficients till 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 be is equal to sixty
- so if we can somehow put this
- in the form
- got a 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 when it's not a right triangle we are really gonna deal with the law of sines or law of cosines
- and a law of sines let me just rewrite it over here just for our reference
- 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 read the law of cosine here
- the law of cosine is c squared
- it should look as a Pythagorean theorem with little adjustment for the fact that is not a right triangle
- So c squared is equal to a squared plus b squared
- minus
- 2 ab
- cosine
- cosine of C of capital C so it's law of sines and law of cosines, so let's 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
- so i have the 2a
- I have 2a
- cosine of c let me write that separately
- so that's 2a
- cosine
- cosine of
- capital c
- and then times
- sine of C over c
- times
- with white 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 let me do this the same thing over here, I have two times
- I'm going to separate
- these guys out
- Actually I wanted to do the sines so let me separate
- I'm gonna separate this guy and in this guy out
- and so I get plus
- 2c
- cosine
- of A
- times sine of capital a over lower case a
- times sine
- of capital a over lower case a and what does this do for me
- Well look at the law of sines right over there I have sine of C over c
- that's that over there
- and then I've sine of A over a and that's that over there capital a over lower case a.
- they're both equal to sign of B over b so we're making progress
- we started introducing b into the equation
- and that's what we actually have information about
- so this
- It could be rewritten a sine of be over b
- so this is the same thing as sine
- of capital b over lower case b
- and this is the same thing a sine
- of capital b over
- lower case 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
- 2c
- it's a lower case c
- 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
- it's the same thing as
- 2a
- 2a
- and I really have a sense of what the next step is so I'll leave a space here
- 2a times cosine of C
- plus this and these are being multiplied I'll have some space there
- plus 2c two lower case c
- times the cosine of A
- and all of this
- all of this times
- the sine
- the sine of B over b and we already know that B is 60 degrees
- so we can evaluate this
- uh... pat
- pretty easily but let's just continue
- let's see if we can somehow put this
- right over here in terms of b
- well if you look over here we have 2a cosine of C
- 2c cosine of A
- it's trying to look
- pretty darn close
- each of these terms look preety 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
- let's see what we could do
- so if you add if you add
- 2ab cosine C to both sides you get 2ab
- 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
- 2ab
- cosine of capital c
- is equal to a squared plus b squared minus
- c squared
- and this is interesting and you know we can switch around the letters later on
- but this looks pretty darn close to the so what if and this looks pretty darn close
- to this except here we're dealing with an a instead of the c, we've just switched
- letters around
- and we could rewrite this
- actually let me rewrite it just for fun
- i could rewrite this over here as
- 2
- 2cb
- not rewrite I can swap the letters
- times the cosine of A
- you see I'm swaping the a(s) and the c(s) is equal to c squared plus b squared minus
- a squared
- there's nothing unique about the sides I can do this with all of the sides
- so here
- It's a big c here, you have an a and b out front
- and then you have tha a squared plus b squared minus the small c squared, if you have the big A then you gonna have
- the cb in the front and you're 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 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(s) 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 the b across here but they were also divided by b
- where which so I'll divide by b, that's an equivalent of multiplying the denominator
- the denominator there not b squared
- that's equivalent of multiplying the denominator there by b. That simple thing is dividing it by b
- We've multiplied it by b divided by b
- That's the same thing
- I'm 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 is that over there
- so it is now in 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 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 with we're just left out with the 2b squared so our whole expression
- has simplified to
- 2b squared sine of B
- sine of capital b
- over
- lower case 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 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 of 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 to the thirty degrees is one-half
- the side opposite to the sixty degrees is square root of three times that so it's a sqaure root of three over two
- You can even use the Pythagorean theorem to figure out
- when you know one of them you could figure out the other one
- So it's the sine of, sinus
- opposite over hypotenuse so square root of three over two over one
- Or just square root of three over two, so this is equal to two
- times
- there's a home stretch, it;s very 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're curious this came from the
- two thousand ten IIT joint entrance examination
- hard to get into engineering and science universities in India
- and they gave this 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 the ideas but anyway I just thought
- it was a pretty pretty neat problem
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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