How do you find the measure of each of the angles of a triangle given the measurements of the sides are 30, 35, 45?

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Answer 1 · 2018-01-03T17:33:20

Angle opposite to 30 #~~41.75# degrees
Angle opposite to 35 #~~59.98# degrees
Angle opposite to 45 #~~87.27# degrees

Use the law of cosines: #c^2=a^2+b^2-2ab*cos(C)#

However, we require angle measures, therefore, we can put the entire equation in terms of #C#:

#c^2=a^2+b^2-2ab*cos(C)#

#=>c^2-a^2-b^2=-2ab*cos(C)#

#=>(c^2-a^2-b^2)/(-2ab)=cos(C)#

#=>(a^2+b^2-c^2)/(2ab)=cos(C)#

#=>C=arccos((a^2+b^2-c^2)/(2ab))#

Angle opposite to 30:

#arccos((35^2+45^2-30^2)/(2*35*45))#

#=arccos(2350/3150)#

#=arccos(47/63)~~ **41.75** #

Angle opposite to 35:

#arccos((30^2+45^2-35^2)/(2*30*45))#

#=arccos(1700/2700)#

#=arccos(17/27)~~ **59.98**#

Angle opposite to 45:

#arccos((30^2+35^2-45^2)/(2*30*35))#

#=arccos(100/2100)#

#=arccos(1/21)~~ **87.27**#