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### Course: Trigonometry > Unit 4

Lesson 6: Challenging trigonometry problems- Trig challenge problem: area of a triangle
- Trig challenge problem: area of a hexagon
- Trig challenge problem: cosine of angle-sum
- Trig challenge problem: arithmetic progression
- Trig challenge problem: maximum value
- Trig challenge problem: multiple constraints
- Trig challenge problem: system of equations

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# Trig challenge problem: arithmetic progression

Sal solves a very complicated algebraic trig problem that appeared as problem 29 in the 2010 IIT JEE Paper I exam. Created by Sal Khan.

## Want to join the conversation?

- hey Sal..instead of replacing the sinA/a and SinC/c by SinB/b...try doing it by Replacing SinA/a by SinC/c and SinC/c by SinA/a...the 'a' and 'c' in the equation are cancelled out and we are left with

2(SinACosC + SinCCosA) = 2{Sin(A+B)}=2{Sin(180-B)}=2SinB(184 votes)- It's really simple -

Using Sine Rule :

Sin A/a = Sin C/c

--> c/a = Sin C/Sin A

--> a/c = Sin A/Sin C

Substitute in Eq -

Sin A and Sin C cancel out leaving,

2(Sin A.Cos C+Sin C.Cos A)

It's in the form of [Sin (A+C)=Sin A.Cos C+Sin C.Cos A]

2Sin(A+C)

Then A+B+C=180

--> A+C=(180-B)

2Sin(180-B)

But Sin(180-B)=Sin B [Allied Angles]

2SinB

=2Sin60

=2.sqrt(3)/2

Ans=sqrt(3)(60 votes)

- Woah. At13:58, when Sal started the drawing the right triangle, I had a moment of ferhoodle, and the cause of that was because my math teacher taught me that in a 30-60-90 triangle, the side opposite the 30 degree angle is 1, the hypotenuse is 2, and the only side left is √3. So when he said it's √3/2, I was a bit confused, especially since I learned this in a sophomore level geometry class, and this is a very reliable source. And when I searched up this predicament, all the finds on the Internet showed both sides of the spectrum, and that did nothing to aid me or even palliate my problem. So, in a nutshell, who is right and why? Thanks!!(7 votes)
- You're both right. You're thinking of it as a 1 - √3 - 2 triangle, and Sal is thinking of it as a 1/2 - √3/2 - 1 triangle. Those two triangles are similar, since your triangle's sides are all twice as large as Sal's, and both of those triangles have angles of 30, 60, and 90 degrees. You'll probably see the same thing with 45-45-90 triangles where some people will say that the sides are 1 - 1 - √2 and others will say that it is √2/2 - √2/2 - 1. Just depends on where you think the unit length should be. Hope this leaves you less ferhoodled!(14 votes)

- I am a tenther.

Can anyone help with this?

Square ABCD has length 13 and points E and F are exterior to the square such that BE=DF=5 and AE=CF=12.

FIND EF^2 ( that is , EF square)(6 votes)- A simple way to solve this would be by extending AE and DF, let them intersect at G; extend EB and FC, let them intersect at H. Then EHFG is actually a square with side = 12 + 5 = 17. Thus the required length is the square of diagonal = 2 * 17^2 = 589 :D(5 votes)

- How is it that sin2C=2sinCcosC?(0 votes)
- sin(A+B)=sin(A).cos(B)+cos(A).sin(B)

sin(A+A)= sin(A).cos(A)+cos(A).sin(A)

sin(2A) =sin(A)[cos(A)+cos(A)]

sin(2A)= sin(A)[2cos(A)]

sin(2A)= 2sin(A).cos(A)(3 votes)

- Can anyone tell me where to find the rest of these problems? Sal starts at #29 but I'm interested in seeing the ones before this.(8 votes)
- You could do this much faster by concluding that if B=60 then A+C=120 (because A+B+C=180). Then using:

sin(A)/a=sin(C)/c

(multiply both sides by c and divide by sin(A))

c/a=sin(C)/sin(A)

also, taking the reciprocal

a/c=sin(A)/sin(C)

And there you go, now you just have to plug this into the beginning equation, that is

(a/c)sin(2C)+(c/a)sin(2A)

Replace a/c and c/a with what we found and use the fact that sin(2x)=2sin(x)cos(x)

[sin(A)/sin(C)]*2sin(C)cos(C)+[sin(C)/sin(A)]*2sin(A)cos(A)

Now cancel the sin(C)with the one in the denominator and also do the same thing for sin(A). Also factor out a 2 and you have

2[sin(A)cos(C)+sin(C)cos(A)]

Now use the fact that sin(x+y)=sin(x)cos(y)+sin(y)cos(x)

2[sin(A+C)]

We know that A+C=120 so sin(A+C)=sin(120)

2sin(120)

And also sin(120)=sin(180-60)=sin(60)=sqrt(3)/2 , and we know that the 2 in the denominator cancels with the one in our expresion so our answer in the end is sqrt(3)(5 votes) - 12:14why is b (denominator) multiplied by b?? why not numerator sin B??(7 votes)
- sky hobo has explained so well...(1 vote)

- Sir

can't we do it in this way ?

given a,b,care in arithmetic progression thus,

2B=A+C --(1)

but in any triangle,A+B+C=180 --(2)

SOLVING (1)&(2)

we get A=60 B=60 C=60

THUS THE TRIANGLE IS EQUILATERAL

therefore all sides are equal a=b=c

putting this values in the given eqn we get it's value = sqroot(3)(3 votes) - This problem can also b solved by using equilateral triangle method.By taking a=b=c=1 and A=B=C=60 degrees.By using this method we can solve it in 1 minute. i think(2 votes)
- That's true. But ideally, by knowing the methods you will be able to solve many other problems too :)(2 votes)

- What is IIT JEE?(1 vote)
- it enginnering entrance exam in india(3 votes)

## Video transcript

- [Instructor] 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 are the lowercase version. So the side opposite to
capital A is lowercase a, the 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 the angles capital
A, capital B, and capital C of the triangle are in
arithmetic progression. Arithmetic progression, very fancy word, but on arithmetic progression is, is a series of numbers that are separated by the same amount. So let me give you some examples. So one, two, three, that's
an arithmetic progression. Two, four, six arithmetic progression. Were separated by two every time. I could do 10, 20, 30, also
an arithmetic progression. These are all arithmetic progression. 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 that tells us, what that tells us about, or maybe tells us maybe it doesn't tell us anything about those angles. So if 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 by two, 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 plus N,
which is equal to A plus 2N. So what does that do? The other thing we know about
three angles in a triangle is that they have to
add up to 180 degrees. So this, this, and this have
to add up to 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 we have
one N and then another 2N, 3A plus 3N is equal to 180 degrees. Or you can divide both sides by three. And you get A plus N is to, A
plus N is equal to 60 degrees. So what does that tell us? Well, A could still be
anything 'cause if N is one, then A is 59. If N is 10, then A is gonna be 50. So it doesn't give us much
information about the angle A. But if you look up here, do
you see an A plus N 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 60 degrees. We're doing pretty well so far. So let's see what we can do at
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 sin of 2C capital C plus c over a sin of two capital A. So let's let me just write it down. So we have, I'll do it in,
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've got a two here, two here, frankly the best things you do is just to 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 two A, let me just rewrite each of
these. So sine of, I should say, sine 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, I always forget the actual names of them, but sine of two times something is two sine of that something times the cosine of that . Times
a 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 the, the
same thing right over here. So sine of two A over here
is going to be two sine of A cosine of A, that's just a standard
trigonometric identity. And we in the trigonometric playlist, we proved that identity. I
think we do it multiple times. And then out in 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 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 to get a B here. And when I think about how do
you get to 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 really going to deal
with the law of sines and 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. And let me just write the
law cosines here just in case it's useful in the future. So
the law of cosine C squared, it's really the Pythagorean
theorem with 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 two a b cosine cosine of C of capital C. So there's law of sines
and law of cosines. 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 has sine of C over c, and this is a 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 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 imma do 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. I'm going to separate
this guy and this guy out. And so I get plus two 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 laws, look at the lot 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 a 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, we ha we've relayed we started introducing B into
the equation and that's where the expression, and that's what we actually
have information about. So this could be rewritten
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, are being multiplied by that two a cosine of capital C times that, and then plus two c, the lowercase 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
this is the same thing as two, a two a and I already have a
sense of what the next step is. I'm 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 lowercase c 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 what we already
know that B is 60 degrees. So we can evaluate this
pretty, pretty easily, but let's just, let's just continue to 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 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 to see what we could do. So if you add, if you add two a b cosine C to both sides, you get two 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 two a b cosine of capital C is equal to a squared plus
b squared minus c squared. And this is interesting. I know we can 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 we're 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 two two c b, not rewrite I can swap the letters
times the cosine of A. 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 C, I can do this with all
of the sides. So here, when you have a big C here,
you have an a and 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're gonna have
the c b 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 in fact, let's multiply this whole
numerator, this whole term by b. 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'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 where at which so I'll divide by b that's the equivalent of
multiplying 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 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 with two b squared. So our whole expression has
simplified to two b squared sine of B sine of capital B
over lowercase b squared. These cancel out. So our whole
expression simplifies to, two 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 two
times the sine of 60 degrees. And if you don't have the
sine of 60 degrees memorize, 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 one. We're dealing with the unit circle. The, the side is 30 degrees. The side opposite. The 30 degrees is one
half the side opposite. The 60 degrees is squared
of three times that so it's squared three over two. 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 to three over two over one, or it's just square to three over two. So this is equal to two, 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 root of three. That's a, that's a pretty neat problem. And just in case, you're curious, this came from the 2010 IIT. IIT's are these hard to get
into engineering and science universities in India, and they give you this exam to
like hundreds of thousands of kids. And, you know, the
top, the top, I don't know, like 2000 actually get into
one of the IITs, but anyway, I just thought it was a
pretty, pretty neat problem.