anonymous
  • anonymous
Find the area of a triangle whose sides have the measures: 6in, 9in, and 11 in.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Heron's formula s=(1/2)(a+b+c)
anonymous
  • anonymous
Im not following you. So the semipermimeter is 13 so what do you do from there?
amistre64
  • amistre64
herons formula can be tedious; but so can the law of cosines...

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amistre64
  • amistre64
11^2 -6^2 -9^2 --------------- = cos(t) -2(6)(9) 9.6.sin(t) --------- = area 2
amistre64
  • amistre64
18sqrt(110) ---------- = abt. 26.97 7
anonymous
  • anonymous
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
amistre64
  • amistre64
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
amistre64
  • amistre64
6sin(t) = h 9 = base hb/2 = area
anonymous
  • anonymous
This is confusing... Im trying to work it
anonymous
  • anonymous
Can you make a picture? I think it would help alot.
amistre64
  • amistre64
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anonymous
  • anonymous
what does 6sin(t) mean?
amistre64
  • amistre64
it means 6 times the sin(angle) sin(angle) = h/6 ; so h = 6 sin(angle)
anonymous
  • anonymous
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
amistre64
  • amistre64
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
anonymous
  • anonymous
but why isn't it c^2 -a^2-b^2+2 ---------------- = cos(C)
anonymous
  • anonymous
ab
amistre64
  • amistre64
becasue "-2ab cosC" is not addition; to solve for cosC you have to divide both sides by "-2ab"

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