



 
Hi Guy, The size of the hole depends on the thickness of the board. I am going to do this where the angle at which the tube meets the board is $d^o$ and the thickness of the board is $t$ inches, and when I'm done set $d = 45^{o}.$ Here is my diagram. Let the distance from $A$ to $B$ be $y$ inches and the distance from $E$ to $A$ be $x$ inches then the diameter of the hole to be drilled is $x + y$ inches. The measure of the angle $CAB$ is $d^o$ and hence \[\tan(d^o) = \frac{t}{y} \mbox { or } y = \frac{t}{\tan(d^o)}.\] Since the sides of the tube are parallel the measure of the angle $DEA$ is $d^{o}.$ The distance from $D$ to $A$ is 5/8 inches and hence \[\sin(d^o) = \frac{5/8}{x} \mbox{ or } x = \frac{5/8}{\sin(d^0)}.\] When $d^o = 45^o$ we have $\tan(d^o) = 1$ and hence $y = t \mbox{ inches}$ and $\sin(d^o) = \large \frac{1}{\sqrt{2}}$ so \[x = \frac{5}{8} \times \sqrt {2} = 0.884 \mbox{ inches.} \mbox{ which is approximately 7/8 of an inch.}\] I hope this helps, 



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