Chapter 13, Trigonometric Ratios and Functions - 13.6 Apply the Law of Cosines - 13.6 Exercises - Skill Practice - Page 893: 41

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We are given: $a,b,c$ Use law of cosines to find, $a^2=b^2+c^2-2bc\cos A\\\cos A=\frac{a^2-b^2-c^2}{-2bc}\\A=\arccos \frac{a^2-b^2-c^2}{-2bc}\\A\approx93.69^\circ$ Find $B=\frac{b^2-a^2-c^2}{-2ac}\\B\approx 33.89^\circ$ Since the sum of angles of a triangle is 180 degrees: $A+B+C=190^\circ\\C=180-A-B=180-93.69-33.89=52.42^\circ$