How to determine sin A in triangle circuit if given all sides of this circuit?work on AB=2,BC=3,AC=4

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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` .

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