Side lengths of inscribed triangle

[Maddie1609]

Homework Statement

The corners A, B and C of a triangle lies on a circle with radius 3. We say the triangle is inscribed in the circle. ∠A is 40° and ∠B is 80°.

Find the length of the sides AB, BC and AC.

Homework Equations

The Attempt at a Solution

I found out the arc AB is 2π, arc BC is 4π/3 and arc AC is 8π/3. Don't know if it's relevant, but I did it nonetheless. The book says nothing about finding the arc or finding the lengths, the question came out of nowhere. It's a shitty book to say the least. Don' really know where to go from here. Is there a constant ratio between length of sides of a central angle and something else?

 

[Noctisdark]

Hint:
1-What's the sum of angle in a triangle ?
2-Does this sound familiar BC/angle(A) = AC/angle(B) = AB/angle(C) = ?
Edit: [= 2R], R is the radius ;)

 

[Maddie1609]

Hint:
1-What's the sum of angle in a triangle ?
2-Does this sound familiar BC/angle(A) = AC/angle(B) = AB/angle(C) = ?

Oh my god, of course, thank you so much!

 

[Noctisdark]

Welcome ;)

 

[LCKurtz]

Does this sound familiar BC/angle(A) = AC/angle(B) = AB/angle(C) = ?

Not to me. It just sounds wrong.

 

[Maddie1609]

I was thinking (s)he meant a/sin A = b / sin B = c/sin C, but I didn't have a single side length so it wasn't much help either way.

Not to me. It just sounds wrong.

 

[LCKurtz]

I was thinking (s)he meant a/sin A = b / sin B = c/sin C, but I didn't have a single side length so it wasn't much help either way.

I supposed that too. So, have you solved the problem?

 

[Maddie1609]

I supposed that too. So, have you solved the problem?

No :-( Any ideas?

 

[LCKurtz]

Hint: Draw the radii to the three vertices. There is a relation between the central angles and the angles of the triangle.

 

[Maddie1609]

Hint: Draw the radii to the three vertices. There is a relation between the central angles and the angles of the triangle.

This is how far I've gotten, but I haven't the slightest clue what to do next. Unless I can somehow find the one other angle in one of the triangles.

 

[LCKurtz]

That's good so far. So for each interior triangle, you have two equal sides and need the third side. The law of Sines wasn't helping you. But there is another law...

 

[Noctisdark]

A/sin(a) = B/sin(b) = C/sin(c) = 2R
Solve for lengh, you hava the radius and the angles,

 

[Maddie1609]

That's good so far. So for each interior triangle, you have two equal sides and need the third side. The law of Sines wasn't helping you. But there is another law...

Cosines :-D Jesus, never would have thought of that, we haven't had any trig in this class. Out of sight, out of mind. Thank you! Is it correct?

 

[Maddie1609]

A/sin(a) = B/sin(b) = C/sin(c) = 2R
Solve for lengh, you hava the radius and the angles,

I'm sorry, but I don't know what you mean.

 

[Noctisdark]

That can only be true if you know two length and one angle, in your case no lenght, I'm going to break the law and tell you the solution,

 

[Maddie1609]

That can only be true if you know two length and one angle, in your case no lenght, I'm going to break the law and tell you the solution,

I solved it above using the law of cosines, but where did the 2r come from? How do you know the ratio is 2×radius of circumscribed circle? :-)

 

[Noctisdark]

You just messed it up and wrongly used the cosine law (al-kashi), just if you read my first post, the sine law is a/sin(A) = b/sin(B) = c/sin(C) = 2R where r is the radius of the circle you are talking about, take it easy .

 

[Maddie1609]

You just messed it up and wrongly used the cosine law (al-kashi), just if you read my first post, the sine law is a/sin(A) = b/sin(B) = c/sin(C) = 2R where r is the radius of the circle you are talking about, take it easy .

Take it easy? Don't know what you're talking about, but okay. What was wrong with it? What does al-kashi mean? Haha. What I'm wondering is how you know the ratio A/sin A = 2r? :-)

 

[Noctisdark]

What i mean is that you over complexed the problem and its wrong , from where did the 3 came of, wrong use of cos law and check any site or your textbook to check the esuation i gave you,

 

[SammyS]

What i mean is that you over complexed the problem and its wrong , from where did the 3 came of, wrong use of cos law and check any site or your textbook to check the equation I gave you,

As Maddie asked: Where did you get the relation, A/sin(A) = 2R ?

It seems LCKurtz didn't consider it to be something commonly known, and I agree with him.

 

[Maddie1609]

What i mean is that you over complexed the problem and its wrong , from where did the 3 came of, wrong use of cos law and check any site or your textbook to check the esuation i gave you,

3 was the radius, but was the equating wrong in and of itself or would you just rather I use a simpler method? I have seen the law of sines before, but I haven't seen that the ratio equals the 2r. Could you explain or tell me what to google? :-)

 

[Noctisdark]

Demonstrate it by yourself, I learned it in basic trig and ever he application of cos law with totally wrong and messed up,

 

[SammyS]

Demonstrate it by yourself, I learned it in basic trig and ever he application of cos law with totally wrong and messed up,

Sure, I can see that it's true for a triangle inscribed in a circle, but it's not something like the Law of Sines and Law of Cosines that I consider to be fundamental.

 

[Maddie1609]

Demonstrate it by yourself, I learned it in basic trig and ever he application of cos law with totally wrong and messed up,

Don't know how I'm supposed to construct a proof for it, but anyhow, I haven't seen it used before in any textbooks I've used so far. Could you explain how my answers were wrong or the use of cosines was wrong?

 

[Noctisdark]

Every thing was wrong about it
http://2000clicks.com/mathhelp/GeometryLawOfSinesProof.aspx

 

[LCKurtz]

I just got back from lunch and have a couple of comments.
@Maddie1609: Your answers are OK but I wouldn't but the value of $a^2$ equal to its square root on the same line.
@Noctisdark: In your post #2 you left out the sines, which is why I said it was wrong.
@SammyS:Yes, I don't consider it commonly known. In fact, I don't remember it even if it is true.

 

[Maddie1609]

Every thing was wrong about it
http://2000clicks.com/mathhelp/GeometryLawOfSinesProof.aspx

I was just looking at that haha, thanks. Like what? I got the same answer you did. Instead of just claiming it was wrong, you could say what exactly was wrong about it.

 

[Maddie1609]

I just got back from lunch and have a couple of comments.
@Maddie1609: Your answers are OK but I wouldn't but the value of $a^2$ equal to its square root on the same line.
@Noctisdark: In your post #2 you left out the sines, which is why I said it was wrong.
@SammyS:Yes, I don't consider it commonly known. In fact, I don't remember it even if it is true.

My bad, ha ha, I was just writing it out swiftly :-) Thanks for your help!

 

[Noctisdark]

You should know it or know how to demonstrate it, but maddie did you explain from where you got the 3, not for me but in the papercheet to give to the teacher, maths should be logic and precise, in fact using sine here is a stright direct application of it not turning around loop and using cos,

 

[Noctisdark]

I was just looking at that haha, thanks. Like what? I got the same answer you did. Instead of just claiming it was wrong, you could say what exactly was wrong about it.

Your work was great but you didn't tell from where you got the 3 and i advice hundred time to consider the simplest and safest path, sorry for making it so loud, and for speaking too much,

 

[Maddie1609]

You should know it or know how to demonstrate it, but maddie did you explain from where you got the 3, not for me but in the papercheet to give to the teacher, maths should be logic and precise, in fact using sine here is a stright direct application of it not turning around loop and using cos,

I couldn't possibly know something I hadn't previously heard of. The fact that the radius is 3 was given in the task. This isn't something I'm handing in, I'm preparing for an exam, so no one but myself sees it :-) That doesn't make it wrong though, haha.