Question

Can someone help with 16,17 and 18 ! thank you !

Answer 1

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:

`tg(x)=\frac{\text{Opposite side}}{Adjacent\text{ side}}`

16)

`\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}`

17)

`\begin{gathered} tg(x)=(8)/(10)=0.8 \\ x=tg^(-1)(0.8) \\ x=39\degree \end{gathered}`

18)

`\begin{gathered} tg(x)=\frac{\text{1}1}{8}=1.375 \\ x=tg^(-1)(1.375) \\ x=54\degree \end{gathered}`