Respuesta :

Solution

- The Cosine law is given below as:

[tex]\begin{gathered} Given\text{ }\triangle ABC,\text{ with sides }a,b,c\text{ and angles }\angle A,\angle B,\angle C\text{ such that} \\ a\text{ is opposite }\angle A \\ b\text{ is opposite }\angle B \\ c\text{ is opposite }\angle C \\ \\ \text{ We have:} \\ a^2=b^2+c^2-2(bc)\cos\angle A \end{gathered}[/tex]

- We can make [tex]\begin{gathered} a^2=b^2+c^2-2bc\cos\angle A \\ \text{ Subtract }b^2\text{ and }c^2\text{ from both sides} \\ \\ a^2-b^2-c^2=-2bc\cos\angle A \\ \\ \text{ Divide both sides by }-2bc \\ \cos\angle A=\frac{a^2-b^2-c^2}{-2bc} \\ \text{ } \\ \text{ Take the cos inverse of both sides} \\ \\ \therefore\angle A=\cos^{-1}(\frac{a^2-b^2-c^2}{-2bc}) \end{gathered}[/tex]

Final Answer

The answer is

[tex]\operatorname{\angle}A=\cos^{-1}(\frac{a^{2}-b^{2}-c^{2}}{-2bc})\text{ \lparen OPTION C\rparen}[/tex]