i hard hard math question

i did a competition today and only one problem i couldn't figure out, it is hard hard hard hard hard, if you can help please do and give your answer and how you did it.

a triangle with lenths 8,10,12 has an angle bisecter that bisecs the angle formed by the 8 and 10 lines, 2 segments are formed on the lenth 12 line, find the lenth of the 2 segments.

Originally posted by Ason Pigg2
i did a competition today and only one problem i couldn't figure out, it is hard hard hard hard hard, if you can help please do and give your answer and how you did it.

a triangle with lenths 8,10,12 has an angle bisecter that bisecs the angle formed by the 8 and 10 lines, 2 segments are formed on the lenth 12 line, find the lenth of the 2 segments.

Solve the angles of the triangle (it's case 2)

http://planetmath.org/encyclopedia/TriangleSolving.html

Then, as the one angle is bisected, you know the values of the two new angles will be half.

Now you have an ASA triangle, which is made up of the 8 (or 10, depending which you pick) and the two known angles. Solve this triangle. It's case 1 in the link above.

This should be in Pozers and Puzzles. I've alerted it, so it might get moved there.

ok, so what if i know the angle degree?it helps me nothing. sin and cos works on right triangles, not this kind.

thanks for the web though

Originally posted by Ason Pigg2
thanks for the web though

Well, I'm not going to spend my time trying to convince you I'm right. I handed you the method you need to solve this, but if you don't believe me, feel free to look elsewhere. You won't have any luck though.

Originally posted by AThousandYoung
Well, I'm not going to spend my time trying to convince you I'm right. I handed you the method you need to solve this, but if you don't believe me, feel free to look elsewhere. You won't have any luck though.

That's a great site thanks AThousandYoung

Originally posted by idioms
That's a great site thanks AThousandYoung

You come up with 55.7, 41.4 and 82.8 degrees for the angles?

Originally posted by sonhouse
You come up with 55.7, 41.4 and 82.8 degrees for the angles?

The question is what are the lengths of the two new lines.

5 and 7

It's actually a "special" triangle with sides in 4:5:6 ratio .. the following page

http://www.mathpuzzle.com/Chebychev.html

Describes the process but essentially this is the bit they want you to show

"If the sides of the triangle (a,b,c) = (1,U[k-1],U[k]), then the angle opposite b will be k times the angle opposite a, a fact easily derived from the Law of Sines."

http://en.wikipedia.org/wiki/Law_of_sines

which will give you the length of the intervals without having to calculate the angles

Here is some information about a Pythagorean triangle.
http://www.mcs.surrey.ac.uk/Personal/R.Knott/Pythag/pythag.html#345

A famous P triangle has it sides: 3,4,5.
The triangle in this thread is 8,10,12 and is not pythagorean.
Neither is the 4,5,6 triangle.

There is no need to know the angles. This is actually pretty simple. (Few days ago I had a test in geometry in a similar theme). If there is a triangle ABC and a bisector AD is drawn from angle A than the bisector rule (I think that's the way it is called in English though I'm not sure) works here => AB/AC = BD/BC. From this you can get an equation system that BD/BC = 8/10 (or 10/8) and BD + BC = 12. By solving this you'll get the result that those two lines are 5 1/3 and 6 2/3. (5.(3) and 6.(6))

Originally posted by kbaumen
There is no need to know the angles. This is actually pretty simple. (Few days ago I had a test in geometry in a similar theme). If there is a triangle ABC and a bisector AD is drawn from angle A than the bisector rule (I think that's the way it is called in English though I'm not sure) works here => AB/AC = BD/BC. From this you can get an equation system th ...[text shortened]... olving this you'll get the result that those two lines are 5 1/3 and 6 2/3. (5.(3) and 6.(6))

I already posted the answer.

🙄

The answer: (16/3) and (20/3), as was quite rightly said earlier.
The method:
- the bisector is perpendicular to the side of 12
- hence there are two right triangles, one with hypotenuse 8, one with hypotenuse 10
- Let the lengths of the division of side 12 be "a" (the larger) and "b"; the angle opposite these sides "x"
- form simultaneous equations:

1) a + b = 12 - i

2) (using sin x = opp/hyp)

sin x = b/8 = a/10, so 4a = 5b - ii

Solving simultaneously gives a = (20/3) and b = (16/3)

I hope this helps and I apologise if I have repeated anything already said.

- the bisector is perpendicular to the side of 12

It's not perpendicular. It would be only if the sidelines of it would be equal. (Triangle ABC, bisector AD, AB=AC)

You're right. How come my method worked then? Luck?