



 
Hi David, In my diagram the triangle ABC has the side CA of length 2 inches and the side BC of length 3 feet 10 inches which is $3 \times 12 + 10 = 46$ inches. Thus \[\tan(CAB) = \frac{46}{2} = 23\] Thus the measure of the angle CAB is given by \[\tan^{1}(23) = 87.5 \mbox{ degrees.}\] Hence you need to cut each of the posts with an angle measuring 87.5 degrees. I hope this helps,  


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