In questions 16, 17 and 18 we have a right triangle and in the three cases you know the measues for 2 sides, except for hipotenuse.
We can use the following relation to solve the problem:
[tex]tg(x)=\frac{\text{Opposite side}}{Adjacent\text{ side}}[/tex]
16)
[tex]\begin{gathered} tg(x)=\frac{\text{1}7}{7}=2.4285 \\ tg(x)=2.4285 \\ x=tg^{-1}(2.4285) \\ x=68\degree \end{gathered}[/tex]
17)
[tex]\begin{gathered} tg(x)=\frac{8}{10}=0.8 \\ x=tg^{-1}(0.8) \\ x=39\degree \end{gathered}[/tex]
18)
[tex]\begin{gathered} tg(x)=\frac{\text{1}1}{8}=1.375 \\ x=tg^{-1}(1.375) \\ x=54\degree \end{gathered}[/tex]
