Problem 535 | Friction on Wedges

$P = 2R \sin (\phi + \frac{1}{2}\alpha)$

$\dfrac{\partial P}{\partial \alpha} = \frac{1}{2}R \cos (\phi + \frac{1}{2}\alpha) = 0$

$\cos (\phi + \frac{1}{2}\alpha) = 0$

$\phi + \frac{1}{2}\alpha = 90^\circ$

$2\phi + \alpha = 180^\circ$

$\alpha = 180^\circ - 2\phi$

$\sin \alpha = \sin (180^\circ - 2\phi)$

$\sin \alpha = \sin 180^\circ \cos 2\phi - \cos 180^\circ \sin 2\phi$

$\sin \alpha = (0)(\cos 2\phi) - (-1)(\sin 2\phi)$

$\sin \alpha = \sin 2\phi$

$\alpha = 2\phi$           answer