Given:
The angle of inclination is,
[tex]\theta=25^{\circ}[/tex]
the frictional force is,
[tex]f=45\text{ N}[/tex]
The mass of the skier is,
[tex]m=62\text{ kg}[/tex]
according to the diagram the frictional force is,
[tex]\begin{gathered} f=\mu mg\cos \theta \\ \mu=\frac{f}{\text{mgcos}\theta} \end{gathered}[/tex]
Substituting the values we get,
[tex]\begin{gathered} \mu=\frac{45}{62\times9.8\times\cos 25^{\circ}} \\ =0.08 \end{gathered}[/tex]
Hence, the coefficient of friction is 0.08.
