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:

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

$=arccos(100/2100)$

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