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