Answer to Question #90786 in Trigonometry for Unknown287159

Question #90786
Solve triangle ABC given AB=16cm ,AC=23cm and BC=21cm
1
Expert's answer
2019-06-14T12:24:00-0400

The Law of Cosines: "\\cos(C)= \\frac{a^2+b^2-c^2}{2ab}"


"\\cos(C)= \\frac{AC^2+BC^2-AB^2}{2\u22c5AC\u22c5BC}=\\frac{23^2+21^2-16^2}{2\u22c523\u22c521}= \\frac{714}{966}\\approx0.74"


"\\angle C=\\arccos{0.74}\\approx42.3^\\circ"


"\\cos(A)= \\frac{AC^2+AB^2-BC^2}{2\u22c5AC\u22c5AB}=\\frac{23^2+16^2-21^2}{2\u22c523\u22c516}= \\frac{344}{736}\\approx0.47"


"\\angle A=\\arccos{0.47}\\approx62.1^\\circ"


"\\angle B=180^\\circ-(42.3^\\circ+62.1^\\circ)=75.6^\\circ"


Answer: "\\angle A=62.1^\\circ, \\angle B=75.6^\\circ, \\angle C=42.3^\\circ."


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