First, notice that since the sum of the internal angles of any triangle must always be equal to 180, then:
[tex]\begin{gathered} \angle N+\angle O+\angle P=180 \\ \Rightarrow119+5+\angle P=180 \\ \Rightarrow\angle P=180-119-5 \\ \Rightarrow\angle P=56 \end{gathered}[/tex]Draw a diagram of the triangle to visualize the situation:
Draw the altitude of the triangle through the vertex N. Let H be the point where the altitude through N intercepts the line OP. From the trigonometric relations, notice the following:
[tex]\begin{gathered} \tan (O)=\frac{HN}{OH} \\ \cos (P)=\frac{HP}{NP} \\ \sin (P)=\frac{HN}{NP} \end{gathered}[/tex]Substitute the known values:
[tex]\begin{gathered} \tan (5)=\frac{h}{OH} \\ \cos (56)=\frac{HP}{3.6\operatorname{cm}} \\ \sin (56)=\frac{h}{3.6\operatorname{cm}} \end{gathered}[/tex]Isolate OH from the first equation, HP from the second and h from the third one:
[tex]\begin{gathered} OH=\frac{h}{\tan (5)} \\ HP=3.6\operatorname{cm}\times\cos (56) \\ h=3.6\operatorname{cm}\times\sin (56) \end{gathered}[/tex]Substitute the expression for h from the third equation into the expression for OH:
[tex]OH=\frac{3.6\operatorname{cm}\times\sin (56)}{\tan (5)}[/tex]Finally, notice that the length of n equals the length of the segment OP, which is the sum of the lengths of the segments OH and HP:
[tex]\begin{gathered} n=OH+HP \\ =3.6\operatorname{cm}\times\frac{\sin(56)}{\tan(5)}+3.6\operatorname{cm}\times\cos (56) \\ =34.1134\ldots cm+2.0131\ldots cm \\ =36.1265\ldots cm \end{gathered}[/tex]Therefore, to the nearest tenth:
[tex]n=36.1\operatorname{cm}[/tex]
