# Side lengths of inscribed triangle

## 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.

## 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

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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 ;)

• 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
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Does this sound familiar BC/angle(A) = AC/angle(B) = AB/angle(C) = ?
Not to me. It just sounds wrong.

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.

• Noctisdark
LCKurtz
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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?

I supposed that too. So, have you solved the problem?
No :-( Any ideas?

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

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. • Noctisdark
LCKurtz
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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,

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?

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• Noctisdark
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.

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

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That can only be true if you know two length and one angle, in your case no lenght, i'm gonna 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? :-)

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 .

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
Staff Emeritus
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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.

• 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? :-)

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

SammyS
Staff Emeritus
Homework Helper
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Demonstrate it by yourself, I learnt 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.

• Noctisdark
Demonstrate it by yourself, I learnt 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