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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*

User RCross
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1 Answer

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11 votes
Answer:

The appropriate answer is:

[B] A = 43.29, B = 30.1°, C = 106.7°

Step-by-step explanation:

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

By the Law of Cosines, we have:


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

SImilarly


B=\cos^(-1)((c^2+a^2-b^2)/(2ac))=30.1082^o

and


C=\cos^(-1)((a^2+b^2-c^2)/(2ab))=106.731^^o

User Broot
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