Please solve Consider a triangle ABC like the one below Supp
Please solve
Solution
by law of cosines
cosA=(b2+c2-a2)/2bc
A=cos-1[(232+692-552)/(2*23*69)]
A=cos-1[0.7136]
A=44.5o
cosB=(a2+c2-b2)/2ac
B=cos-1[(552+692-232)/(2*55*69)]
B=cos-1[0.9561]
B=17o
cosC=(a2+b2-c2)/2ab
C=cos-1[(552+232-692)/(2*55*23)]
C=cos-1[-0.4771]
C=118.5o
