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anonymous

  • 5 years ago

Find the area of a triangle whose sides have the measures: 6in, 9in, and 11 in.

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  1. anonymous
    • 5 years ago
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    Heron's formula s=(1/2)(a+b+c)

  2. anonymous
    • 5 years ago
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    Im not following you. So the semipermimeter is 13 so what do you do from there?

  3. amistre64
    • 5 years ago
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    herons formula can be tedious; but so can the law of cosines...

  4. amistre64
    • 5 years ago
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    11^2 -6^2 -9^2 --------------- = cos(t) -2(6)(9) 9.6.sin(t) --------- = area 2

  5. amistre64
    • 5 years ago
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    18sqrt(110) ---------- = abt. 26.97 7

  6. anonymous
    • 5 years ago
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    Im lost. Law of cosines is c^2 = a^2 + b^2 -2ab cos y right? And I know it can be manipulated but Im still lost

  7. amistre64
    • 5 years ago
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    that is the law of cosines yes; and i used it to find the angle between the legs 6 and 9 so that i could use the sin of that angle to find the area

  8. amistre64
    • 5 years ago
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    6sin(t) = h 9 = base hb/2 = area

  9. anonymous
    • 5 years ago
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    This is confusing... Im trying to work it

  10. anonymous
    • 5 years ago
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    Can you make a picture? I think it would help alot.

  11. amistre64
    • 5 years ago
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  12. anonymous
    • 5 years ago
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    what does 6sin(t) mean?

  13. amistre64
    • 5 years ago
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    it means 6 times the sin(angle) sin(angle) = h/6 ; so h = 6 sin(angle)

  14. anonymous
    • 5 years ago
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    Oka I got that. So how are you supposed to find t then? Can you work out the theorom because I think that's what I'm having trouble with

  15. amistre64
    • 5 years ago
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    i used the law of cosine and modified it to find t c^2 = a^2 + b^2 -2ab cos(C) c^2 -a^2-b^2 ------------ = cos(C) -2ab the cosine inverse of that left side = t

  16. anonymous
    • 5 years ago
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    but why isn't it c^2 -a^2-b^2+2 ---------------- = cos(C)

  17. anonymous
    • 5 years ago
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    ab

  18. amistre64
    • 5 years ago
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    becasue "-2ab cosC" is not addition; to solve for cosC you have to divide both sides by "-2ab"

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