Answer to Question #90786 in Trigonometry for Unknown287159

Question

Solve triangle ABC given AB=16cm ,AC=23cm and BC=21cm

Answer 1

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

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

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

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

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

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

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