Answer to Question #13
See the analysis
[tex]Petermine\ the\ opposite\ side\\ to\ the\ angle\ and\ the\ adjacent\\ Side[/tex]
[tex]tan\propto=\frac{opposite}{adjacent}[/tex]
Answer to Question #11
Answer: see the analysis.
[tex]Solution:\\ The\ sine\ of\ an\ angle\ is\ the\ ratio\ of\ Opposite\ to\ Hypotenuse.\\ So,\ we\ can\ use\ the\ formula\ \sin{\theta}=\frac{Opposite}{Hypotenuse}to\ calculate\ the\ sine\ of\\ an\ angle\ in\ a\ right\ triangle.[/tex]
Answer to Question that asked "How would your results change if you used Angle B Instead of Angle A?"
Ans No Change
[tex]Ans\ No\ change\\ After\ changing\ A\ to\ B\ no\ change\ will\\ be\ there[/tex]
Answer to Question #12
the ansurer is
cose C = Base/Hypitenuse = BC/AC
[tex]Solution:\\ cosine\ of\ angle\ is\ the\\ Natis\ of\ base\ to\ the\\ lypotenuse\ of\ the\ friangle\\ cos\ C=\frac{BC}{AC}[/tex]
I hope this helps you!
:)
