We want the shorter diagonal BD, 
but first we must calculate the longer diagonal, 
the green line AC below:



We use the law of cosines on ΔABC since this is the SAS case:

AC² = AB² + BC² - 2·AC·BC·cos(∠ABC)

AC² = 1.2² + 1.6² - 2(1.2)(1.6)cos(140°)

 AC = 2.634693656 m

We use the law of sines to find ∠BAC





Cross multiply:

(2.634693656)sin(∠BAC) = (1.6)sin(140°)

             sin(∠BAC) = 

                  ∠BAC = 22.97645649°

Now that we have ∠BAC we draw half the shorter diagonal BE, 
which is perpendicular to the longer diagonal AC, in red:



Since ΔABE is a right triangle:

 =  = sin(BAC)

BE = AB·sin(BAC)

BE = 1.2·sin(22.97645649°)

BE = 0.4684234191 m

So the entire shorter diagonal, BD, 



is twice the length of BE, so

BD = 2·BE = 2(0.4684234191) = 0.9368468381 m 

Rounded to tenths:  0.9 m

Edwin