Question 1006182
Draw it helps.  Here, side b is opposite angle B and is line AC; side c is BC, and side a is AB
b^2=a^2+c^2-2ac*cosB
45=25+10-2(sqrt (10)*5*cos B
10/10=-sqrt (10) cos B
sqrt(10)/10=-cos B
cos B= -sqrt(10)/10
=108.43 degrees
a^2=b^2+c^2-2bc cos A
25=45+10-2(sqrt450)) cos A
-30=-2(21.21) cos A
=45 degrees.
as a check, do 
c^2=a^2+b^2-2ab cos C
10=25+45-2(5)(sqrt(45) cos C
-60= -2 *5 sqrt(45) cos C
6=sqrt 45 cos C
cos C=0.8944
=26.6 degrees
With rounding, essentially 180 degrees.