Hi Joanna,

You use the terms horizontal and vertical so I am assuming the the angles $CDA$ and $DAB$ in my rough sketch are right angles.

I drew the line $BE$ parallel to $DA$ and hence angle $CEB$ is a right angle and hence we can use Pythagoras Theorem on the triangle $CEB.$ thus

\[|BC|^2 = |CE|^2 + |EB|^2 = (2560 - 1850)^2 + 1750^2 \;mm^2\]

and hence

\[|BC| = \sqrt{3566600} = 1888.54 \;\; mm\]

Penny