7) The angle asked is ajacent to the leg given and we also have the hypotenuse. So we can use cossine:
[tex]\begin{gathered} \cos x=\frac{8}{18}=\frac{4}{9} \\ x=\arccos (\frac{4}{9})=64\degree \end{gathered}[/tex]
8) Here we want the hypotenus given an angle an its opposite leg. So we can use sine:
[tex]\begin{gathered} \sin (65\degree)=\frac{10}{x} \\ x=\frac{10}{\sin (65\degree)}=11.0 \end{gathered}[/tex]
9) We want the leg which is opposite of a given angle and we have the hypotenuse. So we can use sine again:
[tex]\begin{gathered} \sin (28\degree)=\frac{x}{15} \\ x=15\cdot\sin (28\degree)=7.0 \end{gathered}[/tex]
