Use the Law of cosines to solve triangle ABC given that a = 15, b = 11, and c = 21.[A] A = 34.2", B = 30.1", C= 115.7"[B] A = 43.29, B = 30.1°, C = 106.7°[C] A= 23.4' B = 10.3*, C=1463"D] A = 423', B = 310*, C= 107.6*

Respuesta :

Answer:

The appropriate answer is:

[B] A = 43.29, B = 30.1°, C = 106.7°

Explanation:

Given that a = 15, b = 11, and c = 21.

By the Law of Cosines, we have:

[tex]\begin{gathered} a^2=b^2+c^2-2bc\cos A \\ A=\cos^{-1}(\frac{b^2+c^2-a^2}{2bc}) \\ \\ A=\cos^{-1}(\frac{11^2+21^2-15^2}{2\times11\times21})=\cos^{-1}(\frac{337}{462})=43.1608^o \end{gathered}[/tex]

SImilarly

[tex]B=\cos^{-1}(\frac{c^2+a^2-b^2}{2ac})=30.1082^o[/tex]

and

[tex]C=\cos^{-1}(\frac{a^2+b^2-c^2}{2ab})=106.731^^o[/tex]