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Solve Triangle FGH. Round to the nearest degree. Picture attached!

Mathematics
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@jim_thompson5910 can you help me please?
sorry for getting to you so late, you can use the law of cosines to solve for each angle

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so for instance, let's say we're solving for angle F (ie finding the measure of angle F) The law of cosines is a^2=b^2+c^2-2bc*cos(A) where 'a', b and c are the sides of the triangle and angle A is opposite side 'a'. To find angle F, we do this... a^2=b^2+c^2-2bc*cos(A) (6.3)^2=(5.6)^2+(5.8)^2-2(5.6)(5.8)*cos(F) 39.69=(5.6)^2+(5.8)^2-2(5.6)(5.8)*cos(F) 39.69=31.36+(5.8)^2-2(5.6)(5.8)*cos(F) 39.69=31.36+33.64-2(5.6)(5.8)*cos(F) 39.69=31.36+33.64-11.2(5.8)*cos(F) 39.69=31.36+33.64-64.96*cos(F) 39.69=65-64.96*cos(F) 39.69-65=-64.96*cos(F) -25.31=-64.96*cos(F) (-25.31)/(-64.96)=cos(F) 0.389624384236453=cos(F) cos(F)=0.389624384236453 F=arccos(0.389624384236453) F=67.0688704863168 So the angle F is roughly 67.0688704863168 degrees
It's a lot I know, but hopefully you can go through it step by step without too much trouble and make sense of it all You can then use this idea to find angles G and H.

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