Solution to Problem 122 Shear Stress

$A_{\text{shear}} = tw \csc \theta$

$A_{\text{shear}} = A \, \csc \theta$
 

Shear force,
$V = P \cos \theta$
 

$V = \tau A_{\text{shear}}$

$P \cos \theta = \tau(A \csc \theta)$

$\tau = \dfrac{P \sin \theta \, \cos \theta}{A}$

$\tau = \dfrac{P (2 \sin \theta \, \cos \theta)}{2A}$

$\tau = \dfrac{P \sin 2\theta}{2A}$           (okay!)