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irspow
Either or can be used with any triangle...I personally find the law of sines is often more compact as in this case we can just say;

(sin115)/18=(sin40)/c  (with no need for b or B as would be needed with the law of cosines...)

c=18(sin40)/(sin115)

c≈12.77


BasketballEinstein

Answer:

[tex]\displaystyle 12,766277792...[/tex]

Explanation:

Solving for Angles

[tex]\displaystyle \frac{sin\angle{C}}{c} = \frac{sin\angle{B}}{b} = \frac{sin\angle{A}}{a}[/tex]

Do not forget to use [tex]\displaystyle arcsin[/tex] or [tex]\displaystyle sin^{-1}[/tex]towards the end, or the result will be thrown off.

Solving for Edges

[tex]\displaystyle \frac{c}{sin\angle{C}} = \frac{b}{sin\angle{B}} = \frac{a}{sin\angle{A}}[/tex]

Well, let us get to work:

[tex]\displaystyle \frac{18}{sin\:115} = \frac{c}{sin\:40} \hookrightarrow \frac{18sin\:40}{sin\:115} = c \\ \\ \boxed{12,766277792... = c}[/tex]

I am joyous to assist you at any time.

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