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triangle side=

a=130mm

b=96.8mm

c=56mm

a)what is the angle abc

Guest Jan 28, 2018
 #1
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We can use the Law of Cosines to solve this

 

We have that

 

96.8^2  = 130^2  + 56^2  - 2(130)(56)cos(ABC)       rearrange as

 

[ 96.8^2 - 130^2 - 56^2 ] / [ -2(130)(56)]  =  cos(ABC)

 

Using the cosine inverse (arccos), we have

 

arccos ( [ 96.8^2 - 130^2 - 56^2 ] / [ -2(130)(56)] )  = ABC ≈  42.9°

 

 

cool cool cool

CPhill  Jan 28, 2018

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