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OpenStudy (vera_ewing):
Write an equation that could be used to find the value of b.
http://curriculum.kcdistancelearning.com/courses/GEOMx-HS-A09/b/assessments/R-LawsofSinesandLawofCosinesQuiz/Geometry_8.4_Quiz_FINAL_20q.png
11 years ago
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OpenStudy (vera_ewing):
@marissalovescats it's b2 = 132 + 122 – 2(13)(12)(cos 14°) right?
11 years ago
OpenStudy (marissalovescats):
Yes except it's cos 41 not 14
11 years ago
OpenStudy (marissalovescats):
And 13^2 is 169 not 132
11 years ago
OpenStudy (vera_ewing):
b2 = 122 + 132 – 2(12)(13)(cos 41°)
b2 = 412 + 132 – 2(41)(13)(cos 12°)
b2 = 132 + 122 – 2(13)(12)(cos 14°)
b2 = 122 + 132 + 2(12)(13)(cos 41°)
11 years ago
OpenStudy (vera_ewing):
which one?
11 years ago
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OpenStudy (marissalovescats):
Are these really your answer choices?
11 years ago
OpenStudy (vera_ewing):
yeah they are...
11 years ago
OpenStudy (vera_ewing):
oh wait sorry hold on...
11 years ago
OpenStudy (vera_ewing):
b^2 = 12^2 + 13^2 – 2(12)(13)(cos 41°)
b^2 = 41^2 + 13^2 – 2(41)(13)(cos 12°)
b^2 = 13^2 + 12^2 – 2(13)(12)(cos 14°)
b^2 = 12^2 + 13^2 + 2(12)(13)(cos 41°)
11 years ago
OpenStudy (marissalovescats):
b^2=a^+c^2-2(a)(c)CosB
b^2=12^2+13^2-2(12)(13)Cos41
b^2=144+169-2(12)(13)Cos41
That what it should be... but it would have to be the 1st one I guess
11 years ago
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OpenStudy (vera_ewing):
b^2 = 12^2 + 13^2 – 2(12)(13)(cos 41°) ?
11 years ago
OpenStudy (marissalovescats):
Yes
11 years ago
OpenStudy (vera_ewing):
thank you! :D
11 years ago
OpenStudy (marissalovescats):
Thats why I was confused I was like 122 and 132??? Where are they getting that from? Lol
11 years ago
OpenStudy (vera_ewing):
yeah haha i didn't notice that lol
11 years ago
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OpenStudy (marissalovescats):
That's okay haha
11 years ago
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