Can you find the lengths of a / b by the overall height of the triangle?
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No. Fix the base segment, that fixes two vertices of the triangle. Imagine a line parallel to the base, at a distance of $800$. Take any point on that line, and make that the third vertex. $a,b$ vary dramatically over that family. Nor is $\frac ab$ constant...there is an isosceles example (for which the ratio is $1$) and a right example (for which it is not).
lulu
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Given the assumption that the triangle pictured is isosceles:
$a=b=\sqrt{\bigl(\frac{1700}{2}\bigr)^2+800^2}=1167.261753$
poetasis
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