SOLUTION
The figure below illustrate the situation
Using sine rule it follows:
[tex]\begin{gathered} \frac{\sin92}{16.37}=\frac{\sin66.3}{a} \\ a=15.0 \end{gathered}[/tex]
Also
[tex]\begin{gathered} \frac{\sin92}{16.37}=\frac{\sin21.7}{b} \\ b=6.06 \end{gathered}[/tex]
The vectors are defined by
[tex]\begin{gathered} P=(a\cos15)i+(-a\sin15)j \\ C=(-b\sin17)i+(-bcos17)j \end{gathered}[/tex]
Substituting the the value of a and b into the vectors
It follows:
[tex]\begin{gathered} P=(15\cos15)i+(-15\sin15)j \\ P=14.49i-3.88j \end{gathered}[/tex]
Also
[tex]\begin{gathered} C=(-6.06\sin17)i+(-6.06\cos17)j \\ C=-1.77i-5.80j \end{gathered}[/tex]
Therefore the vectors are:
[tex]\begin{gathered} P=14.49\imaginaryI-3.88j \\ C=-1.77\imaginaryI-5.80j \end{gathered}[/tex]
