[tex]S\frac{O}{H}C \frac{A}{H} T\frac{O}{A}[/tex] or [tex]Sin\frac{opposite}{hypotenuse}Cos \frac{adjacent}{hypotenuse} Tan\frac{opposite}{adjacent}[/tex]
Let's say you are at point B(of the picture), the adjacent side is BC, the opposite side is AC(across from point B), and the hypotenuse is AB. The hypotenuse is the longest side.
Let's say you are at point A, the adjacent side is AC, the opposite side is BC(across from point A), the hypotenuse is AB.
To find BC, we can do:
[tex]cos B = \frac{adjacent}{hypotenuse}[/tex] You plug in the ∠B into B.
[tex]cos 51 = \frac{BC}{14}[/tex] Multiply 14 on both sides
[tex]cos 51(14) = BC[/tex]
8.8 = BC
