Given that
[tex]\angle\text{BKD}=70^o\text{ and }\angle\text{DKC}=20^o[/tex]
We know that
[tex]\angle BKC=\angle BKD+\angle DKC[/tex]
[tex]\text{ Substitute }\angle\text{BKD}=70^o\text{ and }\angle\text{DKC}=20^o,\text{ we get}[/tex]
[tex]\angle BKC=70^o+20^o[/tex]
Adding 70 and 20, we get
[tex]\angle BKC=90^o[/tex]
Hence the angle BKC is 90 degrees.
