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geometry inequality triangles help if you can

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Two sides of an obtuse triangle are 16 and 21. How many possible lengths are there for the third side, if it is a positive integer?

i know that in an obtuse triangle

a^2+b^2<c^2
a+b>c
but i tried plugging in and it wont work so couldyou guys help me?

btw merry christmas and happy new year

  • geometry inequality triangles help if you can - ,

    First let's establish the domain for the third side

    let the third side be c
    c+16>21 ----> c > 5
    c+21 > 16 ----> c > -5 , obviously
    21 + 16 > c ---> c < 37

    so to be a triangle in the third side must be
    5 < c < 37

    Now to be obtuse
    c^2> 21^2+16^2
    c^2 > 697
    c > 26

    could one of the other angles be obtuse ?
    21^2 >c^2 + 16^2
    c^2 < 185
    c < 14

    16^ > c^2 + 21^2
    c^2 < a negative, which is not possible

    So for the angle opposite the third side to be obtuse,
    26 < c < 37

    For the angle opposite the side 21 to be obtuse,
    5 < c < 14

  • geometry inequality triangles help if you can - ,

    thanks a lot the answer is 18 right?

    merry christmas and happy new year to y

  • geometry inequality triangles help if you can - ,

    No, 18 does not work.

    Notice my two possible solutions, .....

    So for the angle opposite the third side to be obtuse,
    26 < c < 37
    So the third side could be 27, 28, 29, 30 31 32 33 34 35 or 36

    For the angle opposite the side 21 to be obtuse,
    5 < c < 14
    so the third side could be 6 7 8 9 10 11 12 or 13

    make a sketch using one of the answers and illustrate that the angle must be obtuse by using the cosine law.

    Also test it with your 18 to show it does not work

  • geometry inequality triangles help if you can - ,

    I meant the total number of possibilities. Thanks anyway!

  • geometry inequality triangles help if you can - ,

    Eighteen is correct. I had the same problem and got it right

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