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6 answers

If each line intersects (shares a point) with each of the other two lines. (Hint, you only have to test one line against the other two, then check and see if the other two intersect in one place.)

This will work in Euclidean or non-Euclidean spaces.

2006-12-05 10:31:09 · answer #1 · answered by Mr. Quark 5 · 0 0

take the three line segments and label them as a, b, and c...

now plug them into the equation:

angle C = cos^-1*(a^2 + b^2 - c^2)/(2*a*b)

this will give you an angle
do the same thing for all the sides

angle B = cos^-1*(a^2 + c^2 - b^2)/(2*a*c)
angle A = cos^-1*(b^2 + c^2 - a^2)/(2*b*c)

if all the angles add up to 180 degrees, then the segments make a triangle...if they do not, or your answers to any of those equations doesn't work, then it is not a triangle

2006-12-05 10:38:01 · answer #2 · answered by Matt M 2 · 0 0

If the three line segments have lengths of a, b, and c, respectively, then they are in a position to form a triangle if and on condition that here 3 circumstances exist: a+b>c a+c>b b+c>a as an occasion, if the three lengths ar 3, 4, and eight, then a triangle is impossible by way of fact 3+4 isn't >8.

2016-10-14 02:28:37 · answer #3 · answered by Anonymous · 0 0

the lines are straight
thier ends are touching
there angles add up to 180 degrees

2006-12-05 10:31:59 · answer #4 · answered by Ralph 7 · 0 0

If any two of the lines add up to more than the third.

2006-12-05 10:29:51 · answer #5 · answered by Elven 3 · 2 0

c=sq rt(a^2+b^2) do it every time.

2006-12-05 10:34:19 · answer #6 · answered by Sophist 7 · 0 0

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