Answer:
Figure 1: Option A : [tex]$ \sqrt{\frac{\textbf{3}}{\textbf{2}}} $[/tex] ; [tex]$ \frac{\textbf{1}}{\textbf{2}} $[/tex] ; [tex]$ \sqrt{\textbf{3}}} $[/tex]
Figure 2: Option B: [tex]$ \frac{\sqrt{\textbf{19}}}{\textbf{10}} $[/tex] ; [tex]$ \frac{\textbf{9}}{\textbf{10}} $[/tex] ; [tex]$ \frac{\sqrt{\textbf{19}}}{\textbf{9}} $[/tex]
Figure 3: Option C: [tex]$ \frac{\textbf{4}}{\textbf{5}} $[/tex] ; [tex]$ \frac{\textbf{3}}{\textbf{5}} $[/tex] ; [tex]$ \frac{\textbf{4}}{\textbf{3}} $[/tex]
Step-by-step explanation:
[tex]$ \textbf{Sin A} \hspace{1mm} \textbf{= } \hspace{1mm} \frac{\textbf{opp}}{\textbf{hyp}} $[/tex]
[tex]$ \textbf{Cos A} \hspace{1mm} {\textbf{=} \hspace{1mm} \frac{\textbf{adj}}{\textbf{hyp}} $[/tex]
[tex]$ \textbf{Tan A} \hspace{1mm} \textbf{=} \hspace{1mm} \frac{\textbf{Sin A}}{\textbf{Cos A}} $[/tex]
We can simply follow these three formulas to solve the problem.
Figure 1:
Sin A = [tex]$ \frac{6\sqrt{3}}{12} $[/tex] = [tex]$ \frac{\sqrt{3}}{2} $[/tex]
Cos A = [tex]$ \frac{6}{12} = \frac{1}{2} $[/tex]
Tan A = [tex]$ \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}} $[/tex] = [tex]$ \sqrt{3} $[/tex]
Figure 2:
Sin A = [tex]$ \frac{2\sqrt{19}}{20} $[/tex] = [tex]$ \frac{\sqrt{19}}{10} $[/tex]
Cos A = [tex]$ \frac{18}{20} = \frac{9}{10} $[/tex]
Tan A = [tex]$ \frac{\frac{\sqrt{19}}{10}}{\frac{10}{9}} $[/tex] = [tex]$ \frac{\sqrt{19}}{9} $[/tex]
Figure 3:
Sin A = [tex]$ \frac{16}{20} $[/tex] = [tex]$ \frac{4}{5} $[/tex]
Cos A = [tex]$ \frac{12}{20} = \frac{3}{5} $[/tex]
Tan A = [tex]$ \frac{\frac{4}{5}}{\frac{3}{5}} $[/tex] = [tex]$ \frac{4}{3} $[/tex]
Hence, the answer.
