Trig (simple stuff for you) | Mathematics | Page 1

cobalt gulch Jan 30, 2025, 8:00 PM

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Ex 5 solve în R the equation if you know one of the solution is x=2pi/3

verbal bladeBOT Jan 30, 2025, 8:00 PM

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cobalt gulch Ex 5 solve în R the equation if you know one of the solution is x=2pi/3

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cobalt gulch Jan 30, 2025, 8:00 PM

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@shut marten

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Also I did ex 4 I need to solve the determinant and got x^3+x^2-10x+8

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Is there a easy way to solve this

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This cubic equation =0

dark vigil Jan 30, 2025, 8:16 PM

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cobalt gulch Ex 5 solve în R the equation if you know one of the solution is x=2pi/3

Well, what would a be if x=2pi/3 is one of the solutions?

dark vigil Jan 30, 2025, 8:17 PM

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cobalt gulch Also I did ex 4 I need to solve the determinant and got x^3+x^2-10x+8

You can try the rational root theorem.

cobalt gulch Jan 30, 2025, 8:17 PM

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dark vigil Well, what would `a` be if x=2pi/3 is one of the solutions?

Everest

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Idk

dark vigil Jan 30, 2025, 8:17 PM

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Just try to plug in x=2pi/3 into the equation.

cobalt gulch Jan 30, 2025, 8:18 PM

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And what I get

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Is there a reason why I do that

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And also does it affect if I have || on coax

shut marten Jan 30, 2025, 8:20 PM

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cobalt gulch Is there a reason why I do that

To find a.

cobalt gulch Jan 30, 2025, 8:20 PM

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Done

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Is -10

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And after

shut marten Jan 30, 2025, 8:20 PM

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Well, remember the formula for cos(2x)?

cobalt gulch Jan 30, 2025, 8:22 PM

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Hmmmmmm

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Let me think let me think

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(Sin2x-1)/2

shut marten Jan 30, 2025, 8:22 PM

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Uh, no.

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I mean the double-angle formula.

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And that isn't correct, anyway.

cobalt gulch Jan 30, 2025, 8:23 PM

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Oh yea double how stupid I’m

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:)))))

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No I don’t know that thing

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Let me check on google

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But there are many formulas

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How do I know which to apply?

shut marten Jan 30, 2025, 8:24 PM

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Well, we want just cosines.

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So, the one that only has a cosine in it.

cobalt gulch Jan 30, 2025, 8:24 PM

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2cos^2-1

shut marten Jan 30, 2025, 8:24 PM

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Also, you should probably learn it, it's one of the basic ones.

shut marten Jan 30, 2025, 8:25 PM

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cobalt gulch 2cos^2-1

Yes. Don't forget the argument, though.

cobalt gulch Jan 30, 2025, 8:25 PM

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shut marten Also, you should probably learn it, it's one of the basic ones.

It’s not like I use it every day

shut marten Jan 30, 2025, 8:25 PM

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cobalt gulch It’s not like I use it every day

That's not the point.

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So, we have:
4|cos(x)| + 3 = -10(2cos(x)^2 - 1)

cobalt gulch Jan 30, 2025, 8:25 PM

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Let me write it

shut marten Jan 30, 2025, 8:26 PM

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Now, for real t we always t^2 = |t|^2, right?

cobalt gulch Jan 30, 2025, 8:26 PM

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Done

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We sub t?

dark vigil Jan 30, 2025, 8:26 PM

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Yeah, that works.

cobalt gulch Jan 30, 2025, 8:27 PM

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Ok and t applys from when ?

shut marten Jan 30, 2025, 8:27 PM

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Yeah! Since cos(x)^2 = |cos(x)|^2, we can just take t = |cos(x)|.

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And we're left with a quadratic equation.

cobalt gulch Jan 30, 2025, 8:27 PM

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Like the domain

shut marten Jan 30, 2025, 8:27 PM

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cobalt gulch Ok and t applys from when ?

Worry about that later.

dark vigil Jan 30, 2025, 8:27 PM

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What is the domain of cosine?

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Oh, alright.

shut marten Jan 30, 2025, 8:28 PM

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dark vigil What is the domain of cosine?

Range, rather.

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Anyway, you can just find t first.

dark vigil Jan 30, 2025, 8:28 PM

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I’m gonna be working it out here, don’t look at the spoiled stuff. ||4t+3=-10(2t^2-1)||

cobalt gulch Jan 30, 2025, 8:28 PM

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576 sort

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Sqrt of 576

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Fast

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Quick math

dark vigil Jan 30, 2025, 8:29 PM

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||4t+3=-20t^2+10||, ||-20t^2-4t+7=0||

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It’s 24.

cobalt gulch Jan 30, 2025, 8:29 PM

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Ok

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Ok

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I got the t

shut marten Jan 30, 2025, 8:30 PM

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cobalt gulch Ok and t applys from when ?

Right.
Now we return to this. Your thoughts?

cobalt gulch Jan 30, 2025, 8:30 PM

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T=-7/10 and t=1/2

dark vigil Jan 30, 2025, 8:30 PM

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||20t^2+4t-7=0||, ||20t^2+14t-10t-7=0||, ||2t(10t+7)-(10t+7)=0||, ||t=1/2, -7/10||

dark vigil Jan 30, 2025, 8:30 PM

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cobalt gulch T=-7/10 and t=1/2

Good.

cobalt gulch Jan 30, 2025, 8:30 PM

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And now

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We do what

shut marten Jan 30, 2025, 8:31 PM

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Well, t = |cos(x)|, so what values can t take?

cobalt gulch Jan 30, 2025, 8:31 PM

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Positive

shut marten Jan 30, 2025, 8:32 PM

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No.

cobalt gulch Jan 30, 2025, 8:32 PM

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Why no

dark vigil Jan 30, 2025, 8:32 PM

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There are certain bounds.

shut marten Jan 30, 2025, 8:32 PM

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Yeah.

dark vigil Jan 30, 2025, 8:32 PM

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Has to do with the range of |cos(x)|.

cobalt gulch Jan 30, 2025, 8:33 PM

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Ok and

shut marten Jan 30, 2025, 8:33 PM

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Well, what is it?

cobalt gulch Jan 30, 2025, 8:33 PM

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We got the solution

shut marten Jan 30, 2025, 8:33 PM

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We haven't solved the problem yet.

cobalt gulch Jan 30, 2025, 8:33 PM

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And we need to see the values that cost can take right?

shut marten Jan 30, 2025, 8:33 PM

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Yes.

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If t = |cos(x)|, what values can t take?

dark vigil Jan 30, 2025, 8:34 PM

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Yeah, but maybe one is extraneous. So we have to rule out if they’re extraneous using the range of |cos(x)|, since that’s how it would work.

cobalt gulch Jan 30, 2025, 8:34 PM

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From -1 to 1

shut marten Jan 30, 2025, 8:34 PM

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That's cos(x).

dark vigil Jan 30, 2025, 8:34 PM

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Not necessarily.

shut marten Jan 30, 2025, 8:34 PM

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What about |cos(x)|?

cobalt gulch Jan 30, 2025, 8:34 PM

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From 0 to 1

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Wait

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It’s right

shut marten Jan 30, 2025, 8:35 PM

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Yeah.

cobalt gulch Jan 30, 2025, 8:35 PM

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0 to 1 lock in

shut marten Jan 30, 2025, 8:35 PM

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More specifically, [0, 1].

cobalt gulch Jan 30, 2025, 8:35 PM

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Yes

shut marten Jan 30, 2025, 8:35 PM

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So, which value of t works?

cobalt gulch Jan 30, 2025, 8:35 PM

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From 0 to 1

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😉

shut marten Jan 30, 2025, 8:35 PM

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No, I get that, but which of the ones that we got?

dark vigil Jan 30, 2025, 8:35 PM

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From the solutions of t earlier.

cobalt gulch Jan 30, 2025, 8:36 PM

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Only 1/2 but my question is why -7/10 wouldn’t work if is modul

shut marten Jan 30, 2025, 8:36 PM

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cobalt gulch Only 1/2 but my question is why -7/10 wouldn’t work if is modul

Because the absolute value can't be negative.

dark vigil Jan 30, 2025, 8:36 PM

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Becuase |x| has only positive ouputs.

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Outputs is key here.

cobalt gulch Jan 30, 2025, 8:36 PM

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Ok

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And now

shut marten Jan 30, 2025, 8:37 PM

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So, now you just need to solve |cos(x)| = 1/2.

cobalt gulch Jan 30, 2025, 8:37 PM

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How

shut marten Jan 30, 2025, 8:37 PM

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You can start by squaring both sides.

cobalt gulch Jan 30, 2025, 8:37 PM

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Wait

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But why |cosx|

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And now cosx

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And not

dark vigil Jan 30, 2025, 8:38 PM

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(Or you can take plus or minus on both sides.)

shut marten Jan 30, 2025, 8:38 PM

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dark vigil (Or you can take plus or minus on both sides.)

Or that.

dark vigil Jan 30, 2025, 8:38 PM

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We set t=|cos(x)| earlier.

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We solved for t.

cobalt gulch Jan 30, 2025, 8:38 PM

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We set cosx=t

dark vigil Jan 30, 2025, 8:38 PM

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Now we want x.

shut marten Jan 30, 2025, 8:38 PM

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The solution will look a bit nicer if you square, though (two series instead of four series).

dark vigil Jan 30, 2025, 8:38 PM

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cobalt gulch We set cosx=t

No.

shut marten Jan 30, 2025, 8:38 PM

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cobalt gulch We set cosx=t

No, we took t = |cos(x)|.

cobalt gulch Jan 30, 2025, 8:39 PM

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And this didn’t affect the cos^2x?

dark vigil Jan 30, 2025, 8:40 PM

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No, since |x|^2=x^2.

cobalt gulch Jan 30, 2025, 8:40 PM

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Ok how to solve

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?

shut marten Jan 30, 2025, 8:42 PM

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As I said, start by squaring both sides.

cobalt gulch Jan 30, 2025, 8:42 PM

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Cos^2x=1/4

shut marten Jan 30, 2025, 8:43 PM

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Right.

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Now you can use the half-angle (or power reduction, whatever you like to call it) formula for cos(x)^2.

cobalt gulch Jan 30, 2025, 8:43 PM

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What formula

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Is this

shut marten Jan 30, 2025, 8:44 PM

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You need to review trigonometric identities.

cobalt gulch Jan 30, 2025, 8:44 PM

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1+cos2x/2

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This one

shut marten Jan 30, 2025, 8:45 PM

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(1 + cos(2x))/2, rather. Right.

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So, (1 + cos(2x))/2 = 1/4.

cobalt gulch Jan 30, 2025, 8:45 PM

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Yes

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Not

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Now

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We multiply by 2?

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@shut marten

shut marten Jan 30, 2025, 8:46 PM

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Yes...

cobalt gulch Jan 30, 2025, 8:46 PM

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Ok and now subtract 1?

shut marten Jan 30, 2025, 8:46 PM

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Yes.

cobalt gulch Jan 30, 2025, 8:47 PM

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And now

shut marten Jan 30, 2025, 8:47 PM

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And now it's an elementary trigonometric equation.

cobalt gulch Jan 30, 2025, 8:47 PM

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Cos2x=-1/2

shut marten Jan 30, 2025, 8:47 PM

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Yes.

cobalt gulch Jan 30, 2025, 8:48 PM

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How to solve

shut marten Jan 30, 2025, 8:48 PM

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Uh...

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You don't know how to solve elementary trigonometric equations?

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You need to review how to do that, then.

cobalt gulch Jan 30, 2025, 8:48 PM

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😦

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Teach me and I will know

shut marten Jan 30, 2025, 8:49 PM

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You should've already learned it if you are solving problems like this.

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If you don't know at all, then learn the topic first.

cobalt gulch Jan 30, 2025, 8:49 PM

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I know but I don’t remember

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Very well

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I see that cos2x=-1/2

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From my equation

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And is 4pi/3

shut marten Jan 30, 2025, 8:51 PM

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You need all solutions.

cobalt gulch Jan 30, 2025, 8:52 PM

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And 2pi/3?

shut marten Jan 30, 2025, 8:53 PM

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You need all solutions.

cobalt gulch Jan 30, 2025, 8:53 PM

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And I don’t see anymore

shut marten Jan 30, 2025, 8:53 PM

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An elementary trigonometric equation has infinitely many solutions.

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Like, sorry, but if you can't see more, then you just haven't learned this topic.

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Or don't remember it at all.

cobalt gulch Jan 30, 2025, 8:54 PM

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I don’t remember

shut marten Jan 30, 2025, 8:54 PM

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In any case, you need to review it.

cobalt gulch Jan 30, 2025, 8:54 PM

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I forgot my main text book in class

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On how to solve

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But I just started to solve again

shut marten Jan 30, 2025, 8:54 PM

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Just look up any school-level algebra textbook.

cobalt gulch Jan 30, 2025, 8:55 PM

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I wrote that thing from October and I frogot till now

cobalt gulch Jan 30, 2025, 8:55 PM

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shut marten Just look up any school-level algebra textbook.

Can I Finnish this first

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I have still one more question

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I don’t want to be unfinished

shut marten Jan 30, 2025, 8:56 PM

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If you are encountering questions like this, you need to be able to solve trigonometric equations.

cobalt gulch Jan 30, 2025, 8:58 PM

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Ok

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How to solve it

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Now

shut marten Jan 30, 2025, 8:58 PM

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Ugh, fine...

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The solutions of cos(x) = a are x = ±arccos(a) + 2πn, n ∈ ℤ.

cobalt gulch Jan 30, 2025, 8:59 PM

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Yes

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I remember something like this

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But in my case I have cos2x

shut marten Jan 30, 2025, 9:00 PM

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Well, that doesn't really matter.

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Express 2x first.

cobalt gulch Jan 30, 2025, 9:01 PM

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Like this?

shut marten Jan 30, 2025, 9:01 PM

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Right. What is arccos(-1/2)?

cobalt gulch Jan 30, 2025, 9:02 PM

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2pi/3

shut marten Jan 30, 2025, 9:02 PM

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Yeah.

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So, 2x = ±2π/3 + 2πn, n ∈ ℤ.

cobalt gulch Jan 30, 2025, 9:02 PM

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I just checked on google don’t worry I didn’t knew that

shut marten Jan 30, 2025, 9:02 PM

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Thus, what is x?

cobalt gulch Jan 30, 2025, 9:03 PM

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Like this?

shut marten Jan 30, 2025, 9:04 PM

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Yeah. Though, you can obviously simplify 2/6 a bit.

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Other than that, yeah, that's the answer.

cobalt gulch Jan 30, 2025, 9:04 PM

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But I remember from book it was something like that

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()U()

shut marten Jan 30, 2025, 9:05 PM

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Intervals come up in inequalities, not equations.

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Well, I mean, they can sometimes come up in equations, too, but that is usually when you have some tricky stuff with absolute values and all that.

cobalt gulch Jan 30, 2025, 9:06 PM

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So like the way we solved is also correct

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?

shut marten Jan 30, 2025, 9:06 PM

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Well, not really "also".

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We just solved it, that's it.

cobalt gulch Jan 30, 2025, 9:07 PM

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This solution can be also positive and negative?

shut marten Jan 30, 2025, 9:07 PM

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Well, obviously.

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Why not?

cobalt gulch Jan 30, 2025, 9:07 PM

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Just check

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Ok so if we would need to make and intervals how we can make it

shut marten Jan 30, 2025, 9:08 PM

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That isn't related to this problem.

cobalt gulch Jan 30, 2025, 9:09 PM

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Yea but this is the solution

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And interval

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An

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I remember very well

shut marten Jan 30, 2025, 9:09 PM

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What are you talking about?

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We found the solution, that's it.

cobalt gulch Jan 30, 2025, 9:09 PM

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The solution

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From the book the solution is an interval

shut marten Jan 30, 2025, 9:09 PM

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Can you show?

cobalt gulch Jan 30, 2025, 9:10 PM

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I can’t right now

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It’s also at school

shut marten Jan 30, 2025, 9:10 PM

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Well, I just don't really understand what intervals you want here.

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The set of solutions is discrete, it doesn't contain any intervals.

cobalt gulch Jan 30, 2025, 9:11 PM

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So there was an interval for what x can take solutions

shut marten Jan 30, 2025, 9:11 PM

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Oh, that!

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Well, yes, some problems include an interval that you need to pick the solutions from.

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Not here, though.

cobalt gulch Jan 30, 2025, 9:11 PM

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Ok

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Why I have op on my name?

shut marten Jan 30, 2025, 9:12 PM

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cobalt gulch Why I have op on my name?

OP means "original poster".

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As in, you're the one who created the post.

cobalt gulch Jan 30, 2025, 9:12 PM

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Oh yea

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Ok now this one is the last

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So we need to find the determinant and put = 0

shut marten Jan 30, 2025, 9:13 PM

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I assume you need to find the values of x for which the matrix is invertible?

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Or not invertible?

shut marten Jan 30, 2025, 9:13 PM

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cobalt gulch So we need to find the determinant and put = 0

In any case, yes.

cobalt gulch Jan 30, 2025, 9:13 PM

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Is inversable

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And I got x^3+x^2-10x+8=0

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How I can solve this

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Like is there a trick or something

shut marten Jan 30, 2025, 9:15 PM

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Try using integer root theorem.

cobalt gulch Jan 30, 2025, 9:15 PM

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Show me

shut marten Jan 30, 2025, 9:15 PM

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If you don't know it, just seach what it is and try it yourself first.

cobalt gulch Jan 30, 2025, 9:16 PM

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You also use this method?

shut marten Jan 30, 2025, 9:17 PM

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Actually, hold on. Are you sure about that determinant?

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Ah, wait.

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You canceled by 2 already, nevermind.

shut marten Jan 30, 2025, 9:18 PM

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cobalt gulch You also use this method?

And yeah, I would.

cobalt gulch Jan 30, 2025, 9:19 PM

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It’s already late here and need to wake up at 6

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I will make it tomorrow

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,to

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,ti

glass locustBOT Jan 30, 2025, 9:19 PM

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The current time for gicu4130 is 11:19 PM (EET) on Thu, 30/01/2025.

cobalt gulch Jan 30, 2025, 9:19 PM

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@shut marten ,ti

shut marten Jan 30, 2025, 9:19 PM

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Just one hour of difference.

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It's 12:19 AM for me now.

cobalt gulch Jan 30, 2025, 9:20 PM

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Why you don’t sleep

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Get some rest :))

shut marten Jan 30, 2025, 9:20 PM

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Nah, I slept for a couple of hours during the day. Not sleepy yet.

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You should sleep, though.

cobalt gulch Jan 30, 2025, 9:22 PM

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Ok man

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Спасибо за помощь

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😌

shut marten Jan 30, 2025, 9:23 PM

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You're welcome!

cobalt gulch Jan 31, 2025, 9:39 AM

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We did something wrong

shut marten Jan 31, 2025, 9:44 AM

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What do you mean?

cobalt gulch Jan 31, 2025, 8:37 PM

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cobalt gulch

The response îs 2pik

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I got pol

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Pik

dark vigil Jan 31, 2025, 8:47 PM

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Hmm.

shut marten Jan 31, 2025, 8:48 PM

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What?

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I don't understand what you're saying, sorry.

dark vigil Jan 31, 2025, 8:48 PM

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Well, their answer is right, but not specific enough?

shut marten Jan 31, 2025, 8:48 PM

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No, like, what's the problem?

dark vigil Jan 31, 2025, 8:49 PM

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shut marten I don't understand what you're saying, sorry.

They (the answer key) got -pi/3+2pik, npi, and 2pi/3+2pik as solutions.

shut marten Jan 31, 2025, 8:49 PM

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dark vigil They (the answer key) got -pi/3+2pik, npi, and 2pi/3+2pik as solutions.

Yeah, same as what we got.

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Just written in a longer way.

dark vigil Jan 31, 2025, 8:49 PM

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dark vigil Well, their answer is right, but not specific enough?

Hence.

shut marten Jan 31, 2025, 8:50 PM

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No, I mean, it's correct. They just obviously did cos(x) = ±1/2 and didn't bother checking whether the two series can be combined into one.

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It's like writing x ∈ {-1, 1} instead of x = ±1.

cobalt gulch Feb 1, 2025, 8:33 AM

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@shut marten

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Hello

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Have some time to talk?

shut marten Feb 1, 2025, 10:14 AM

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cobalt gulch <@290136354233384960>

Hello.
Sure.

cobalt gulch Feb 1, 2025, 10:14 AM

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Problem 8

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For me i wasted 1-2 hours to make it but cant

shut marten Feb 1, 2025, 10:15 AM

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Can you translate it?

dark vigil Feb 1, 2025, 10:16 AM

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,rcw

glass locustBOT Feb 1, 2025, 10:16 AM

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dark vigil Feb 1, 2025, 10:17 AM

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Oh, geometry.

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Glad I just finished that chapter.

dark vigil Feb 1, 2025, 10:24 AM

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cobalt gulch

My first attempt would be to consider cyclic quadrilaterals.

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I translated it to the best of my ability without a software so I might be wrong if I made a few assumptions.

cobalt gulch Feb 1, 2025, 11:12 AM

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shut marten Can you translate it?

No

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I dont know how to Translate romb

cobalt gulch Feb 1, 2025, 11:31 AM

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help

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@shut marten

dark vigil Feb 1, 2025, 11:34 AM

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dark vigil My first attempt would be to consider cyclic quadrilaterals.

?

cobalt gulch Feb 1, 2025, 11:34 AM

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you think i know what is cyclic quadrialitias?

shut marten Feb 1, 2025, 11:34 AM

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cobalt gulch <@290136354233384960>

I can't help you, as I don't know the language.

cobalt gulch Feb 1, 2025, 11:35 AM

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wait let me translate

dark vigil Feb 1, 2025, 11:35 AM

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cobalt gulch you think i know what is cyclic quadrialitias?

It’s simply a quadrilateral that is contained inside of a circle.

cobalt gulch Feb 1, 2025, 11:35 AM

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no

dark vigil Feb 1, 2025, 11:35 AM

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If not that, you can note that the diagonals of a rhombus intersect perpendicularly.

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And that they bisect their respective angles.

cobalt gulch Feb 1, 2025, 11:38 AM

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We consider to have a diamond ABCD with m(<ABC)=120*, AC∩BD = [O]. Let it be a point M in the middle of the side [BC], AM∩BD= [E] and OE=2cm. Find the area of the diamond ABCD.

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@shut marten

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@dark vigil

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this is how i would translate

shut marten Feb 1, 2025, 11:39 AM

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No need to ping all the time.

dark vigil Feb 1, 2025, 11:39 AM

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cobalt gulch We consider to have a diamond ABCD with m(<ABC)=120*, AC∩BD = [O]. Let it be a ...

Oh, so M is a midpoint, as I expected. This makes it much easier.

cobalt gulch Feb 1, 2025, 11:39 AM

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ok

shut marten Feb 1, 2025, 11:39 AM

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By diamond, I assume a rhombus is meant.

cobalt gulch Feb 1, 2025, 11:39 AM

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yes

dark vigil Feb 1, 2025, 11:40 AM

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Try drawing it out first. I had my drawing but I am leaving for someplace so I don’t have it anymore.

cobalt gulch Feb 1, 2025, 11:40 AM

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ok wait

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This îs al that i ca-n make

dark vigil Feb 1, 2025, 11:47 AM

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Yeah, that’s what I had.

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Let me re-analyze it.

shut marten Feb 1, 2025, 11:49 AM

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We can ||drop a perpendicular from M onto AC||, which allows us to ||find BO by using some similar triangles||. And it's easy after that.

cobalt gulch Feb 1, 2025, 11:49 AM

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lord if you make this problem u for real genius

cobalt gulch Feb 1, 2025, 11:49 AM

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shut marten We can ||drop a perpendicular from M onto AC||, which allows us to ||find BO by ...

how

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my teacher used other stuffs here but i frogot how he solved

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he even sat and think like 5 min how to solve

dark vigil Feb 1, 2025, 11:50 AM

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You’ve used similar triangles before, right?

cobalt gulch Feb 1, 2025, 11:50 AM

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the only thing i remember he used ao=xsqrt3over2

cobalt gulch Feb 1, 2025, 11:50 AM

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dark vigil You’ve used similar triangles before, right?

not really

dark vigil Feb 1, 2025, 11:51 AM

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Ah.

dark vigil Feb 1, 2025, 11:51 AM

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cobalt gulch the only thing i remember he used ao=xsqrt3over2

And how do you think that came up?

shut marten Feb 1, 2025, 11:52 AM

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cobalt gulch my teacher used other stuffs here but i frogot how he solved

Well, I mean, there's probably more than one way to solve it.

dark vigil Feb 1, 2025, 11:52 AM

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I remember writing down some trig bash.

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I’ll see if I recall it.

cobalt gulch Feb 1, 2025, 11:52 AM

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dark vigil And how do you think that came up?

u can see its a equirateral triangle

shut marten Feb 1, 2025, 11:52 AM

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dark vigil I remember writing down some trig bash.

Remember? You've seen this problem before?

shut marten Feb 1, 2025, 11:53 AM

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cobalt gulch u can see its a equirateral triangle

I don't want to use that.

cobalt gulch Feb 1, 2025, 11:53 AM

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why

shut marten Feb 1, 2025, 11:53 AM

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We are given an angle α and a distance x, that's enough for me.

cobalt gulch Feb 1, 2025, 11:53 AM

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ok lets do your method after the teacher method

dark vigil Feb 1, 2025, 11:53 AM

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shut marten Remember? You've seen this problem before?

No, I was working on it an hour ago.

cobalt gulch Feb 1, 2025, 11:53 AM

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explain please your aproach

shut marten Feb 1, 2025, 11:53 AM

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dark vigil No, I was working on it an hour ago.

Ohh, ok.

cobalt gulch Feb 1, 2025, 11:54 AM

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approach or aproach?

dark vigil Feb 1, 2025, 11:55 AM

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Yeah, you can just cheese it and ||make x any value that makes sense||.

dark vigil Feb 1, 2025, 11:55 AM

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cobalt gulch approach or aproach?

Two p.

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sotrue

cobalt gulch Feb 1, 2025, 11:56 AM

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dark vigil Yeah, you can just cheese it and ||make x any value that makes sense||.

no

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@shut marten you here?

shut marten Feb 1, 2025, 11:56 AM

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Yes.

cobalt gulch Feb 1, 2025, 11:56 AM

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lets see you vs my teacher

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he is also ex soviet

shut marten Feb 1, 2025, 11:57 AM

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Try it yourself first.

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It's a geometry probem, I'm not that interested in it.

cobalt gulch Feb 1, 2025, 11:57 AM

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i tried like 1-2 hours

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and cant get the responde

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response

shut marten Feb 1, 2025, 11:57 AM

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shut marten We can ||drop a perpendicular from M onto AC||, which allows us to ||find BO by ...

Well, I've written the basic idea of my approach here.

cobalt gulch Feb 1, 2025, 12:04 PM

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how do you know is similar

dark vigil Feb 1, 2025, 12:05 PM

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The first thing to do is to recognize which triangles are similar.

cobalt gulch Feb 1, 2025, 12:06 PM

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how

dark vigil Feb 1, 2025, 12:07 PM

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Do you know the rules for similarity?

cobalt gulch Feb 1, 2025, 12:07 PM

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i found something else

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what if we go from M to DC

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and we make another side from A to N

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so we have an isoscel triangle

dark vigil Feb 1, 2025, 12:08 PM

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Same concept, just with the isosceles instead of the right triangles.

cobalt gulch Feb 1, 2025, 12:08 PM

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Like this

#

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is this a way?

dark vigil Feb 1, 2025, 12:09 PM

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Yeah, same thing.

cobalt gulch Feb 1, 2025, 12:09 PM

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ok how to solve

dark vigil Feb 1, 2025, 12:09 PM

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Similar triangles.

cobalt gulch Feb 1, 2025, 12:09 PM

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how

dark vigil Feb 1, 2025, 12:10 PM

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Do you know the rules for similarity?

cobalt gulch Feb 1, 2025, 12:10 PM

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no

dark vigil Feb 1, 2025, 12:10 PM

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Okay.

cobalt gulch Feb 1, 2025, 12:11 PM

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Teach me and I will know

dark vigil Feb 1, 2025, 12:12 PM

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Three rules:
• AAA similarity (3 angles in 2 triangles are congruent)
• SAS similarity (Two sides and an angle between the two sides are the same)
• AAS similarity (two angles and a side between are the same)

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So what you can do here is identify which is similar using these rules, then apply a ratio.

cobalt gulch Feb 1, 2025, 12:14 PM

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Aaa

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But I don’t see it

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@shut marten

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And what we can find with your method

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If we create a side till AC

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We can see let’s say similarity with abo and let’s call Ann

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Amj

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You don’t get anything

dark vigil Feb 1, 2025, 12:28 PM

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@cobalt gulch Let’s say we have 2 similar triangles ABC and DEF; then we can create a ratio AB/DE=BC/EF.

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Sorry, I was in an area with no service.

cobalt gulch Feb 1, 2025, 12:29 PM

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Ok

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And in my case

dark vigil Feb 1, 2025, 12:30 PM

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Note that triangles inside each other are similar. By what property is this true?

cobalt gulch Feb 1, 2025, 12:32 PM

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Ok they are similar

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But I can’t get anything from them

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I don’t like this

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I want to solve the problem

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And after that to ask question

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I waste so much

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Much

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Much

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Time by doing this

dark vigil Feb 1, 2025, 12:46 PM

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cobalt gulch But I can’t get anything from them

Ratios.

cobalt gulch Feb 1, 2025, 12:46 PM

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Show me

dark vigil Feb 1, 2025, 12:47 PM

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Find the line segment from M to AC first, then use that to find BO.

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Since M is the midpoint, what does that line segment from M to AC do to triangle BOC?

cobalt gulch Feb 1, 2025, 12:49 PM

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Right triangle

cobalt gulch Feb 1, 2025, 1:43 PM

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Ok I give up on this

cobalt gulch Feb 2, 2025, 10:13 AM

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If anyone want to solve I will be happy

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@glacial dune

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Yo bro

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How is going

glacial dune Feb 2, 2025, 10:14 AM

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holy shit

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idk what the question is

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nor am i in any mood to help

dark vigil Feb 2, 2025, 1:28 PM

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cobalt gulch Right triangle

Good. Sorry, I was somewhere else and didn't have good service.

shut marten Feb 2, 2025, 9:57 PM

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dark vigil Good. Sorry, I was somewhere else and didn't have good service.

Oh, this is so trippy...

dark vigil Feb 2, 2025, 10:02 PM

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shut marten Oh, this is so trippy...

cobalt gulch Feb 3, 2025, 9:05 AM

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dark vigil Good. Sorry, I was somewhere else and didn't have good service.

And now

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Brother just show me you way to solve

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We waste s-o much time

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After that I can ask questions

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@shut marten show the way sir

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Пж

shut marten Feb 3, 2025, 9:07 AM

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Have you at least tried the approach I proposed above?

cobalt gulch Feb 3, 2025, 9:08 AM

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But I don’t see what you see

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I stare at that rhombus like a mannequin

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To see the idea but I can’t

shut marten Feb 3, 2025, 9:09 AM

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Have you drawn the perpendicular?

cobalt gulch Feb 3, 2025, 9:15 AM

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Urs

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Yes

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ABO~AMK

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What i get

cobalt gulch Feb 3, 2025, 6:29 PM

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Thanks guy 4 days for a problem

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The best server

dark vigil Feb 3, 2025, 6:29 PM

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cobalt gulch Thanks guy 4 days for a problem

I mean, if you don’t follow, why don’t you try and learn it?

cobalt gulch Feb 3, 2025, 6:29 PM

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How i cam learn?

dark vigil Feb 3, 2025, 6:30 PM

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Visual patterns. And the rules I gave earlier.

cobalt gulch Feb 3, 2025, 6:30 PM

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Im dumb to use it

dark vigil Feb 3, 2025, 6:30 PM

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cobalt gulch But I can’t get anything from them

Use ratios.

cobalt gulch Feb 3, 2025, 6:30 PM

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What’s next

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I want to understand the logic

dark vigil Feb 3, 2025, 6:31 PM

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I know.

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What ratios can we form in similar triangles?

cobalt gulch Feb 3, 2025, 6:31 PM

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Tell me

dark vigil Feb 3, 2025, 6:31 PM

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I can’t.

cobalt gulch Feb 3, 2025, 6:31 PM

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We make proportion

dark vigil Feb 3, 2025, 6:31 PM

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I’m asking you, what ratios can we form from this triangle.

cobalt gulch Feb 3, 2025, 6:32 PM

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Give an example with ratio

dark vigil Feb 3, 2025, 6:32 PM

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Let me see if I can draw something.

cobalt gulch Feb 3, 2025, 6:33 PM

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The area of ABO is 2 times bigger than AMK?

dark vigil Feb 3, 2025, 6:35 PM

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What are O and K?

cobalt gulch Feb 3, 2025, 7:11 PM

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O the center of rhombus and K the line down for mm

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From M

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Do the problem

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After We talk

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You need 40 min to make a draw

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Nu e

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Nu e

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Nice

dark vigil Feb 3, 2025, 7:49 PM

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cobalt gulch You need 40 min to make a draw

I have classes. Calm down.

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I’ll try to figure this out when I get home.

cobalt gulch Feb 3, 2025, 7:50 PM

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Ping me when you solve it

dark vigil Feb 3, 2025, 7:50 PM

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I already did.

cobalt gulch Feb 3, 2025, 7:50 PM

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,time @dark vigil

glass locustBOT Feb 3, 2025, 7:50 PM

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The current time for exiled_hype is 02:50 PM (EST) on Mon, 03/02/2025.
gicu4130 is 7 hours ahead, at 09:50 PM (EET) on Mon, 03/02/2025.

dark vigil Feb 3, 2025, 7:54 PM

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cobalt gulch Ping me when you solve it

Look, I’ve already solved it, but I don’t have paper on me and it might take 2 more hours for me to get home at worst.

cobalt gulch Feb 3, 2025, 7:54 PM

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What response you got?

dark vigil Feb 3, 2025, 7:55 PM

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I can’t give you the answer.

cobalt gulch Feb 4, 2025, 6:42 AM

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I aleardy tot the answer i want to see if you got it

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I aleardy made the problem

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But other way

dark vigil Feb 4, 2025, 12:10 PM

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cobalt gulch I aleardy tot the answer i want to see if you got it

Ok.

glacial dune Feb 5, 2025, 10:00 AM

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Best server ever

obtuse robinBOT Feb 6, 2025, 10:00 AM

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@cobalt gulch

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#Trig (simple stuff for you)