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find angle b given that c equals 58, a equals 49, and angle a equals 56 using law of sines

Guest Mar 5, 2017
edited by Guest  Mar 5, 2017
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Law of Sines:

\(\frac{\sin A}{a} = \frac{sin B}{b} = \frac{\sin C}{c}\)

 

Plug in what you know.

\(\frac{\sin56}{49} = \frac{\sin C}{58} \\~\\ \frac{58\sin56}{49} = \sin C \\~\\ \arcsin({\frac{58\sin56}{49}}) = C \\~\\ C \approx 78.905^{\circ}\)

 

So since we know that there are 180º in every triangle,

 

180 ≈ 56 + 78.905 + B

B ≈ 180 - 56 - 78.905

B ≈ 45.095º

hectictar  Mar 5, 2017

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