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Mathematics
OpenStudy (anonymous):

Which set of three segments can form a triangle? 12, 30, 5 32, 19, 22 29, 35, 6 16, 31, 14

OpenStudy (anonymous):

is it triangle?

OpenStudy (anonymous):

which one forms a triangle?

OpenStudy (anonymous):

:( i need posotive answers -.-

OpenStudy (anonymous):

isnt it supposed to be equilateral triangle and not just a triangle

OpenStudy (anonymous):

not likely...none of the choices has two equal lengths

OpenStudy (anonymous):

Its IMPOSSIBLE :(

OpenStudy (anonymous):

my best bet to go with is the third one

OpenStudy (anonymous):

its second one 32, 19, 22

OpenStudy (anonymous):

no. way. O.O yu figured it out! r yu sure?

OpenStudy (anonymous):

I think I know of a way to solve this one. The law of cosines (valid for any triangle) says:\[c^2=a^2+b^2-2abcos \theta\]Now, lets try the first set and see what we get:\[900=169-2(5)(12)\cos \theta=169-120\cos \theta\]So,\[-6.09=\cos \theta\]Since cosine of an angle lies between -1 and 1, we have a contradiction. Thus 12, 30, 5 cannot form a triangle. Try this for the next three sets and see what you get.

OpenStudy (anonymous):

i did the same equation

OpenStudy (anonymous):

It's the second one. It works in law of cosines and it is correct.

OpenStudy (anonymous):

the sum of the lengths of any two line segments must be greater than the length of the third line segment. 32+19=51>22 32+22=54>19 19+22=41>32 so,32, 19, 22

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