A triangular lot has sides of 215m, 185m, and 125m. Find the measures of the angles at its corners

Answer:The measures of the angles at its corners are [tex]59.1\°,35.4\°,85.5\°[/tex]Step-by-step explanation:see the attached figure to better understand the problemstep 1Find the measure of angle AApplying the law of cosines[tex]185^{2}= 215^{2}+125^{2}-2(215)(125)cos(A)[/tex][tex]2(215)(125)cos(A)= 215^{2}+125^{2}-185^{2}[/tex][tex]cos(A)= [215^{2}+125^{2}-185^{2}]/(2(215)(125))[/tex][tex]cos(A)=0.513953[/tex][tex]A=arccos(0.513953)=59.1\°[/tex]step 2Find the measure of angle BApplying the law of cosines[tex]125^{2}= 215^{2}+185^{2}-2(215)(185)cos(B)[/tex][tex]2(215)(185)cos(B)= 215^{2}+185^{2}-125^{2}[/tex][tex]cos(B)= [215^{2}+185^{2}-125^{2}]/(2(215)(185))[/tex][tex]cos(B)=0.81489[/tex][tex]B=arccos(0.81489)=35.4\°[/tex]step 3Find the measure of angle CApplying the law of cosines[tex]215^{2}= 125^{2}+185^{2}-2(125)(185)cos(C)[/tex][tex]2(125)(185)cos(C)= 125^{2}+185^{2}-215^{2}[/tex][tex]cos(C)= [125^{2}+185^{2}-215^{2}]/(2(125)(185))[/tex][tex]cos(C)=0.0784[/tex][tex]C=arccos(0.0784)=85.5\°[/tex]