Question 1068730
Assuming we are talking about triangle ABC we have sides:

a which is opposite angle A
b which is opposite angle B
c which is opposite angle C

We have angle A that is in between sides b and c
We have angle B that is in between sides a and c
We have angle C that is in between sides a and b.

So since we have angle B that is in between sides a and c (and we know all three of these values), this lets us know that we can use the law of cosines.

So

{{{b^2 = a^2 + c^2 - 2*a*c*cos(B)}}}

{{{b^2 = 6^2 + 25^2 - 2*6*25*cos(14.3)}}}

{{{b^2 = 370.3}}}

{{{b = 19.243}}}

We need to find angle A. So since we know side a, side b and angle B, we can use law of sines.

{{{sin(A)/a = sin(B)/b}}}

{{{sin(A) =  a*sin(B)/b}}}

{{{sin(A) = 6*sin(14.3)/19.243}}}

{{{sin(A) = 0.07701}}}

{{{highlight(A = 4.42)}}}