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# triangle...?

If the sides of a triangle are 6.45 , 5.32 , and 7.33 long - how are the angles?

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- 1 decade agoFavorite Answer
law of cosines:

let side A be opposite angle a, B opp. angle b, C opp. angle c

and let A=6.45; B=5.32, C=7.33.

then by the law of cosines,

C^2 = 7.33^2 = 6.45^2 + 5.32^2 - 2*6.45*5.32*cos(c)

==> c = arccos([7.33^2 - 6.45^2 - 5.32^2]/[2*6.45*5.32])= 1.809

by law of sines,

sin(1.809)/7.33 = sin(a)/6.45 ==> a = arcsin(6.45sin(1.809)/7.33)= 1.0257

and sin(1.809)/7.33 = sin(b)/5.32 ==> b=arcsin(5.32sin(1.809)/7.33) = 0.783

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