Problem 329 | Equilibrium of Force System

$\Sigma M_O = 0$

$100 R \cos \alpha = 200 R \cos \beta$

$\cos \alpha = 2 \cos \beta$
 

 

From the figure
$\alpha + \beta = 90^\circ$

$\alpha = 90^\circ - \beta$
 

Thus,
$\cos (90^\circ - \beta) = 2 \cos \beta$

$\sin \beta = 2 \cos \beta$

$\dfrac{\sin \beta}{\cos \beta} = 2$

$\tan \beta = 2$

$\beta = 63.43^\circ$           answer
 

$\alpha = 90^\circ - 63.43^\circ$

$\alpha = 26.57^\circ$           answer