Minimising the pulling force on a sledge rope modelled by a trigonometric quotient

A child pulls a loaded sledge across level snow by means of a rope. The rope is inclined at \(\theta\) radians to the horizontal (see the diagram).

The smallest tension, \(F\) newtons, in the rope that keeps the sledge moving is modelled by

\(F(\theta)=\dfrac{k}{\cos\theta+\mu\sin\theta},\quad 0\leq\theta\leq\dfrac{\pi}{2},\)

where \(k\) and \(\mu\) are positive constants. When the rope is horizontal, \(F=240\).

1. Show that \(k=240\).[2]

The tension is also \(240\) newtons when \(\theta=\dfrac{\pi}{3}\).

2. Show that \(\mu=\dfrac{\sqrt3}{3}\).[4]

The tension takes its least value when the rope is inclined at \(\beta\) radians to the horizontal, for some angle \(\beta\) with \(0<\beta<\dfrac{\pi}{2}\).

3. 1. Show that \(\dfrac{\mathrm{d}F}{\mathrm{d}\theta}=\dfrac{720\left(3\sin\theta-\sqrt3\cos\theta\right)}{\left(3\cos\theta+\sqrt3\sin\theta\right)^2}\).
   2. Hence show that \(\beta=\dfrac{\pi}{6}\). Find also the exact value of \(F(\beta)\).[9]