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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: area of a triangle

Sal solves a very complicated geometrical trig problem that appeared as problem 11 in the 2003 AIME II exam. Created by Sal Khan.

## Want to join the conversation?

- The confusing part for me is that the question says point D is on the same side of AB as C. Is that badly worded or is it just me?(39 votes)
- It's not just you. That is where I started to get lost, too. I watched the video a second time and suddenly it all made sense! They were just trying to tell us that the point D is inside the triangle ABC.(31 votes)

- At12:06, how did you get 2 (25/2) ^2 (1 + sine theta) from 2 (25/2) ^2 + 2 (25/2)^2 sine theta?(19 votes)
- The expression 2(25/2)^2 + 2(25/2)^2sin theta has two terms. Both of these terms has a common factor 2(25/2)^2. Sal "pulled" this factor out of both terms.

Let's make a substitution so it might be easier to see: let x = 2(25/2)^2. Then the expression becomes x + xsin theta. We pull the x out to become x(1 + sin theta). This is simple factoring out, or Sal sometimes calls it undistributing. We can redistribute the x to both terms by multiplying through to get x + xsin theta.(22 votes)

- The Point "D" should not be positioned inside the triangle. 15+15 = 30 which is only 1 less than 24+7 which makes you wonder how it could be inside the triangle. Note that 5*11^1/2/2 is greater than 8. Compare with the triangle base of 7. Did I get something wrong here?(14 votes)
- I was just wondering this exact same thing. It's physically impossible to have that triangle inside the given triangle. I even took it to graph paper and used a compass and it's impossible to have point D where he drew it.(6 votes)

- At5:23, how can he deduce that it will be a right triangle and that line DM is perpendicular to line AB?(11 votes)
- He knows it is a right triangle, because if you draw a line from the
`midpoint`

of an isosceles triangle to its apex, it is by definition a right triangle. That midpoint means that you form two triangles such that the angles on either side of that midpoint add up to 180 degrees, plus all their corresponding sides and angles are the same. Two equal angles that add up to 180 degrees have to be 90 degrees. Ta da!

Draw a couple of isosceles triangles (two identical sides) and you can remind yourself that this is true. Some of Sal's great videos in the geometry topics can review proofs of this for you. It makes life easier when working with triangles to be confident of that fact.(6 votes)

- So genius can do this in 4 minutes, right?(please reply)(5 votes)
- I've been in similar situations so here are some things I'd like to share (hopefully this isn't too much for a reply to an "off-topic" question) :

1) Trigonometry is like learning the alphabet compared to the famous problems that Euler, Ramanujan and Gauss solved. Some kids learn the alphabet faster and know more words, but that doesn't determine if they would go on to become great novelists.

2) At this stage it's more about practice and less about inherent skill. Many university students with iqs of about 120 would be able to solve this problem in 4 mins, simply because they have practiced more. It's too early to be equating mathematical success with your genius.

3) I think if someone only knew the basics of trig (sin,cos,tan) and tried to attempt this problem, they would take much time to answer it because the number of ways to think about approaching this problem with only that information is huge. Imagine Ramanujan with the mathematical information and exposure of the average 15 year old but leave himself with his own IQ. I think he would take way more than 4 mins.(10 votes)

- Right at the end, when asked to add m+n+p, I would have used sqrt11 to represent n, as it was derived as such, not just n: why did Sal use just 11 rather than Sqrt11?(5 votes)
- it says in the intro that n has to be an integer, n's root would not be an integer.(5 votes)

- 10:40, I begin to loose out, what is that law of cosines that sal is referring to?(4 votes)
- At10:50, its cos(x)=sin(90-(x)) cos(theta+90), so why does it become sin(90-(theta-90))? Shouldnt it be sin(90-(theta PLUS 90)) instead?(4 votes)
- If you look at it this way, it should help: cos(theta+90)=sin(90-(theta+90)). Which is the same as sin(90-theta-90).(4 votes)

- I haven't finished the last unit but I have no idea how to solve this. After finishing the last unit would i have an idea of how to solve this?(4 votes)
- You would have an idea, it takes time and practice though. Repetitions of manipulating the different angle identities help you see patterns in these problems, where you would otherwise spend hours working out the identities manually. Keep up the practice and remember to roll back a unit if you're stuck or confused. Vince Lombardi brought the Green Bay Packers to victory by starting at the fundamentals and drilling them into those NFL players. Never forget the hidden power of mastering the fundamentals.

Then, one step at a time, you will find problems like these as simple as Sal does.(2 votes)

- At5:21，I really can't get why line DM is perpendicular to line AB(3 votes)
- AD = BD and AM = BM, so both D and M are equal distances from A and B, which makes DM perpendicular to AB.(4 votes)

## Video transcript

Triangle ABC is a right
triangle with AC is equal to 7, BC is equal to 24, and
a right angle at C. So let's just try to draw that. And let's draw the right angle. Maybe we'll have to
use some coordinates. So let's draw the right
angle at the origin. So let's say that this
is C right over here. AC is equal to 7. So A, we'll put over here. This distance over
here is going to be 7. And then we have the
hypotenuse of our triangle. And then this can be B. And
this distance over here-- they tell us that BC is equal to 24. All right. Now, point M is
the midpoint of AB. So point M is right over here. Let me do that in
a different color. Point M is the midpoint of AB. So this distance is
equal to that distance. And D is on the same side of
AB as C. So C is on this side, on I guess you'd call it the
bottom left-hand side of AB, so that AD is equal
to BD is equal to 15. So D is going to be
someplace over here. It's going to be equal
distance between A and B. Obviously, all of the points
equal distant between A and B are going to sit on a line
that looks something like that. This is the midpoint of AB. So D is going to
sit right over here. And it's 15 away
from both A and B. So it would look
something like that. That distance over there is 15. And then this distance over
here is also going to be 15. Given that the area of
triangle CDM-- this is D. So CDM is the triangle that
they want to think about. Given that the area of triangle
CDM may be expressed as m times the square root of n over
p, where m, n, and p are positive integers and m
and p are relatively prime. So that just means
you can't simplify it. And n is not divisible by
the square of any prime. So you've simplified the
radical as much as possible-- find m plus n plus p. So we essentially
need to find the area of this green triangle,
of CDM over here. And so let's see what we
can do to figure it out. So we could try some
of the coordinates for some of these points. So this point
right over here, A, is going to be-- its x
value is going to be 7. I could draw the coordinate,
just so you see what I'm doing. That could be the x-axis. And then this side
is on the y-axis. So the coordinate for
A would be 7 comma 0. Coordinate for C
would be 0 comma 0. And the coordinate
for B would be 0, 24. So the coordinate for M would
just be the average of B and A. So the coordinate for M, the
average of 0 and 7 is 7/2. And the coordinate
for the y-coordinate, the average of 24 and 0, is 12. That's fair enough. Now let's see what we can
figure out about the sides. So we know this is
a right triangle. So our gut reaction is always
to use the Pythagorean theorem. We know this side and that side. So if we wanted
to figure out AB, we could just say
that-- we just know that 24 squared plus 7 squared
is equal to AB squared. And 24 squared is 576, plus
49 is equal to AB squared. And let's see-- 576 plus 49. If it was plus 50, it would
get us to 626, but it's 1 less than that. So it's 625. So 625 is equal to AB squared. So AB is going to equal 25. So this distance, the distance
of this big hypotenuse here, is 25. Or half of the
distance, this is going to be 25/2, from B to M. From M
to A is also going to be 25/2. Now, the other thing we know
is that M right over here, the triangle CMA is
an isosceles triangle. How do we know that? Well, M, if you look
at its x-coordinate-- its x-coordinate is directly
in between the x-coordinates for C and A. 7/2--
it's the average. This is 7. This is 0. Its M-coordinate is
right over there. It's directly above the
midpoint of this base over here. So this is going to be
an isosceles triangle. It's symmetric. You could flip
the triangle over. So this length-- and
this seems useful because this is kind of the base
of the triangle we care about. We could kind of view
this as the base of CDM-- this is also going to be 25/2. It's an isosceles triangle. This is going to be the
same as that because we're symmetric around
this right over here. So let's see. We know one side
of this triangle. Let's see if we could figure
out this side over here. It seems pretty straightforward. This is going to be a right
triangle right over here, because this line
from D to M is going to be perpendicular to AB. All of the points that are
equidistant between A and B are going to be on a line
that is perpendicular to AB. So this is going to
be a right triangle. So we can figure out DM using
the Pythagorean theorem again. We get 25/2 squared
plus DM squared is going to be
equal to 15 squared. It's going to equal to the
hypotenuse of this triangle. So it's going to equal 225. And so what do we get? We get DM squared is
equal to 225 minus 625/4. So let's put this over 4. So 225 with 4 as the denominator
is the same thing as 900/4. And in the last video, I
actually mistakenly said 900/4 was 125, obviously a
boneheaded mistake. But the numbers
show up again here. So 225 is obviously 900/4. And then we're going to
subtract from that 625/4. And this is equal
to-- let's see. In our numerator, we
have 900 minus 625 would be 300 minus 20. It's going to be 275/4. And so DM is going to be equal
to the square root of this. So it's equal to the
square root of 275/4. 275 is 25 times 11, because
25 times 12 would be 300. So 25 times 11 over 4. So this is going to be equal
to 5 square roots of 11 over 2. So DM is 5 square
roots of 11 over 2. So what we need to
do here is figure out the height of this
triangle right over here. If we could figure out the
height of that triangle, we're pretty much done. 1/2 the base times the height,
but we don't know-- let's see, we could figure that if we
know the law of cosines. We could do
something over there. If we know the sine of
this angle right over here, the sine of that
angle is going to be this height over the
side we just figured out. So if we could figure out
the sine of that angle, then we'd be done, or we'll
be very close to being done. But it's not any obvious way. But one thing we can do,
if you look at this bigger triangle over here--
so let me highlight it. So if you look at triangle BMC. Let me draw triangle BMC. Well, I'll just draw it on
this drawing right over here. I don't want to
get too overloaded. But triangle BMC
right over here-- we know this side is 25/2. We know this side
over here is 25/2. We know this side
over here is 24. And so what we want
to do is figure out the sine of this angle right
here, the sine of theta, of angle CMD. Now, that's hard to do. But what we can
do is use the law of cosines of theta plus 90. So let me redraw our triangle. So triangle BCM I
could redraw like this. It's actually an
isosceles triangle. So we have BCM. And what is this
angle right over here? It's going to be our theta that
we care about plus 90 degrees. So this angle right here
is theta plus 90 degrees. And this over here is 24. This is 25/2, and this is 25/2. And so using this, we could
use the law of cosines to figure out what
theta actually is here. So let's do that. We're going to use a little
bit of trig identities, but law of cosines. So we get the opposite
to the angle squared, so we get 24 squared. 24 squared is equal to 25/2
squared, plus 25/2 squared, minus 2 times 25/2 times 25/2
times the cosine of this angle, times-- let me scroll
over-- the cosine of theta plus 90 degrees. Now, you're saying,
hey, Sal, you know, this has it in terms of cosine
of theta plus 90 degrees. How can we figure out
the sine of theta? That's what we
actually care about, to figure out the
area of this triangle, or actually to figure out
the height of this triangle. And to do that, you just
have to make the realization. We know the trig identity
that the cosine of theta is equal-- I won't use
theta because I don't want to overload theta-- cosine of x
is equal to sine of 90 minus x. So the cosine of
theta plus 90 degrees is going to be equal to
the sine of-- let me put it in parentheses-- 90
minus whatever is here. 90 minus theta minus 90,
which is equal to-- the 90's cancel out-- sine
of negative theta. And we know sine
of negative theta is equal to the
negative sine of theta. So this over here
simplifies to-- this is the negative sine of theta. So we could write
the sine of theta here and then put the
negative out here. And this becomes a positive. So what does this simplify to? We have 24 squared,
which is 576. 576 is equal to-- let's see. I won't skip any steps here. So we have 25 squared
plus 25 squared. This is 2 times 25/2
squared plus 2 times-- this is 25/2 squared
again-- times sine of theta. Now we just have to
solve for sine of theta. So this is going to
be equal to-- well, this is 576 is equal to
2 times 25/2 squared. I'm just factoring that out. 1 plus sine of theta. Or we can just divide both sides
of the equation by this here. Actually, let me
just simplify it. This thing over here is
625/4, but then we're going to multiply that by 2. So this thing over
here is 625/2. So let's divide both
sides of this by 625/2. And we get 576
times 2 over 625-- just multiplying both sides
by the inverse-- is equal to. When you multiply both sides
by the inverse [? like this, ?] it actually cancels out-- is
equal to 1 plus sine of theta. Or we just subtract
1 from both sides. We get sine of theta is
equal to 576 times 2. Let's see, 76 times
2 is 152, plus 1,000. So it's 1,152/625. That's that part there, minus 1. But instead of 1,
let's say 625/625. So minus 625. And this is equal to-- I'll
have to do a little math on the side-- so
1,152 minus 625. You get a 12 there. This becomes a 4. 12 minus 5 is 7. 4 minus 2 is 2. 11 minus 6 is 5. So this is equal to 527/625. Now, you might not realize it,
but we're in the home stretch now. Let me draw this CDM
triangle, the one that is really the focus
of the problem. And let me draw it a
little bit differently. Let me draw it
slightly differently. So now we know some very
interesting things about CDM. So this is C, this
is D, and this is M. We know this
side over here. We figured out it's
5/2 square root of 11. That's the length of DM. We also know that CM
right over here is 25/2. We also know that sine of
theta here is equal to 527/625. That's what we just figured out. And we can use that to figure
out the height of this triangle because we know that sine is the
opposite over the hypotenuse, if we draw a right
triangle right here. So sine of theta,
which is 527/625, is equal to the opposite,
is equal to the height of this triangle,
over the hypotenuse, over 5/2 square roots of 11. So we can multiply both sides
of this by 5/2 square roots 11. And we get the height
of the triangle is equal to 527/625 times
5/2 square roots of 11. And let's see, we can divide--
625 divided by 5 is 125. So you get a 1 here. And you have a 125 over here. So this is equal
to 527 square roots 11 over 125 times
2, which is 250. That's the height. Now, what's the area
of the triangle? It's 1/2 base times height. Area is equal to 1/2 base,
which is 25/2, times the height, which we just figured out is
527 square roots of 11 over 250. Let's see. Divide the numerator by 25. Divide the denominator by 25. You get a 10 there. So this is equal to 527 square
roots of 11 over 2 times 2 times 10, which is 40. So that's our area. And the whole
problem-- they didn't want us to find the area. They wanted us to find
m plus n plus p, so they want us to find essentially
527 plus 11 plus 40. So 527 plus 11 is-- let me
make sure I do this-- 538, and then plus 40 is 578. And we're done.