Calculate: 80^2 = 42.2^2 + 60^2 – 2(42.2)(60)cosA
2007-01-11 03:23:51 · 5 answers · asked by thomasgraham880 1 in Science & Mathematics ➔ Mathematics
Let a=80
Let b = 42.2
Let c = 60 so that
a^2 = b^2 + c^2 -2bccosA and
-2bccosA = a^2-b^2-c^2
cosA = (b^2+c^2-a^2)/(2bc)
A = arccos{(b^2+c^2-a^2)/(2bc)}
2007-01-11 03:31:47 · answer #1 · answered by kellenraid 6 · 1⤊ 1⤋
80^2 = 42.2^2 + 60^2 – 2(42.2)(60)cosA
6400= 1780.84 +3600 -5064cosA
1019.16 = 5064 cosA
cosA= 0.291256
A = arccos .291256 = 78.39 degrees
2007-01-11 03:43:31 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
80² = 42.2² + 60² - 2(42.2) (60) cos A
6400 = 1780.84 + 3600 - 5064 cos A
6400 = 5380.84 - 5064 cos A
5064 cos A = (- 6400 + 5380.84)
cos A = -1019.16/5064
A = Inv cos (- 0.201255924)
A =101.61º
2007-01-11 04:10:07 · answer #3 · answered by edison c d 4 · 0⤊ 0⤋
A= Inv cos (-0.201255924)
A= 101.61
2007-01-11 03:41:41 · answer #4 · answered by ENA 2 · 0⤊ 1⤋
A = 101.61 degrees
2007-01-11 03:47:21 · answer #5 · answered by ag_iitkgp 7 · 0⤊ 0⤋