Physics - NLM Question with Solution | TestHub

The given situation can be shown as As the plane rises 5 in 12 $$\therefore \sin \theta =\frac{5}{12}$$
and $\cos \theta =\sqrt{1-{\sin }^2\theta }=\sqrt{1+{\left(\frac{5}{12}\right)}^2}$ $$=\frac{\sqrt{119}}{12}$$ So, the coefficient of friction, $$\mu =\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{5}{12}\times \frac{12}{\sqrt{119}}=\frac{5}{\sqrt{119}}=0.46$$