Question 1019495
<pre>
{{{drawing(400,4400/37,-40,1070,-40,290,

triangle(0,0,694,0,334.827817,250.2285615),
locate(330,0,694),locate(660,290,694),
locate(95,155,418),locate(890,155,418),locate(520,155,602),

locate(0,0,A),locate(694,0,B),locate(1028,290,C),locate(335,290,D),
triangle(1028.827817,250.2285615,694,0,334.827817,250.2285615) )}}}

{{{drawing(400,4400/37,-40,1070,-40,290,

locate(330,0,694),locate(660,290,694),
locate(95,155,418),locate(890,155,418),locate(520,155,"???"),
triangle(0,0,1028.827817,250.2285615,334.827817,250.2285615),
locate(0,0,A),locate(694,0,B),locate(1028,290,C),locate(335,290,D),
triangle(0,0,694,0,1028.827817,250.2285615) )}}}

We will be using the law of cosines.
Since the adjacent interior angles of a parallogram are
supplementary, cos(&#8736;ABC) = -cos(&#8736;DAB), so

Using the law of cosines on &#916;ABD

       BD² =  AD² +  AB² - 2&#8729; AD&#8729; AB&#8729;cos(&#8736;DAB)
(1)   602² = 418² + 694² - 2&#8729;418&#8729;694&#8729;cos(&#8736;DAB)  

Using the law of cosines on &#916;ABC

       AC² =  BC² +  AB² - 2&#8729; BC &#8729;AB&#8729;cos(&#8736;ABC)
       AC² = 418² + 694² - 2&#8729;418&#8729;694&#8729;[-cos(&#8736;DAB)]
(2)    AC² = 418² + 694² + 2&#8729;418&#8729;694&#8729;cos(&#8736;DAB)

Adding equations (1) and (2)

(1)   602² = 418² + 694² - 2&#8729;418&#8729;694&#8729;cos(&#8736;DAB)
(2)    AC² = 418² + 694² + 2&#8729;418&#8729;694&#8729;cos(&#8736;DAB)

602² + AC² = 2&#8729;418² + 2&#8729;694² 
       AC² = 2&#8729;418² + 2&#8729;694² - 602²
       AC² = 2&#8729;418² + 2&#8729;694² - 602²
       AC² = 349448 + 963272 - 362484
       AC² = 950316
        AC = &#8730;<span style="text-decoration: overline">950316</span>
        AC = 974.8415256 ft.   

Edwin</pre>