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)=a2+b2c22ab\cos(C)= \frac{a^2+b^2-c^2}{2ab}


cos(C)=AC2+BC2AB22ACBC=232+21216222321=7149660.74\cos(C)= \frac{AC^2+BC^2-AB^2}{2⋅AC⋅BC}=\frac{23^2+21^2-16^2}{2⋅23⋅21}= \frac{714}{966}\approx0.74


C=arccos0.7442.3\angle C=\arccos{0.74}\approx42.3^\circ


cos(A)=AC2+AB2BC22ACAB=232+16221222316=3447360.47\cos(A)= \frac{AC^2+AB^2-BC^2}{2⋅AC⋅AB}=\frac{23^2+16^2-21^2}{2⋅23⋅16}= \frac{344}{736}\approx0.47


A=arccos0.4762.1\angle A=\arccos{0.47}\approx62.1^\circ


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


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


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