Take point d on AB so that AD=AC,

Let DH⊥AC, H be vertical foot, BC=3k, AB=5k,

Then AD=AC=4k,

In △ADH, ∠ AHD = 90 and sin ∠ A =.

∴DH=ADsin∠A= 12/5k，AH= 16/5k。

Then in △CDH, ch = AC-ah = 4/5k, and CD = 4 √ 10/5k.

So in delta △ACD, AD = AC = 4k.

According to the definition of antithesis, sadA= CD/AD=√ 10/5, that is, Sadα = √ 10/5.