Answer
$d = \frac{b}{2}~(cot~\frac{\alpha}{2}+~cot~\frac{\beta}{2})$
Work Step by Step
Let $d_1$ be the distance from point P to the center of the bar. We can find an expression for $d_1$:
$\frac{b/2}{d_1} = tan(\frac{\alpha}{2})$
$d_1 = \frac{b}{2~tan(\frac{\alpha}{2})}$
$d_1 = \frac{b~cot(\frac{\alpha}{2})}{2}$
Let $d_2$ be the distance from point Q to the center of the bar. We can find an expression for $d_2$:
$\frac{b/2}{d_2} = tan(\frac{\beta}{2})$
$d_2 = \frac{b}{2~tan(\frac{\beta}{2})}$
$d_2 = \frac{b~cot(\frac{\beta}{2})}{2}$
The total distance $d$ from point P to point Q is $d_1+d_2$. We can find an expression for $d$:
$d = d_1+d_2$
$d = \frac{b~cot(\frac{\alpha}{2})}{2}+\frac{b~cot(\frac{\beta}{2})}{2}$
$d = \frac{b}{2}~(cot~\frac{\alpha}{2}+~cot~\frac{\beta}{2})$
