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How to determine sin A in triangle circuit if given all sides of this circuit?work on...
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You should make use of the law of cosines such that:
`BC^2 = AB^2 + AC^2 - 2AB*AC*cos hat A`
`BC^2- AB^2- AC^2 = - 2AB*AC*cos hat A`
`cos hat A = (AB^2 + AC^2 - BC^2)/(2AB*AC)`
`cos hat A = (2^2 + 4^2 - 3^2)/(2*2*4)`
`cos hat A = (4 +16 - 9)/(16)`
`cos hat A = 11/16`
You need to use the fundamental formula of trigonometry such that:
`sin hat A = sqrt(1 - cos^2 hat A)`
`sin hat A = sqrt(1 - 121/256)`
`sin hat A = sqrt((256-121)/256)`
`sin hat A = sqrt135/16 => sin hatA = 3sqrt15/16`
Hence, evaluating the sine of the angle A in triangle circuit yields `sin hat A = 3sqrt15/16` .
Posted by sciencesolve on July 29, 2012 at 3:19 PM (Answer #1)
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