Question 933855
check your side {{{a}}}; with given {{{a=0.025}}} is impossible to get real solution for angle {{{A}}} using law of cos:

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

{{{(0.025)^2=10^2+(9.978)^2-(10)(9.978)*cos(A)}}}

{{{0.000625=100+99.560484-99.78*cos(A)}}}

{{{0.000625=199.560484-99.78*cos(A)}}}

{{{99.78*cos(A)=199.560484-0.000625}}}

{{{99.78*cos(A)=199.559859}}}

{{{cos(A)=199.559859/99.78}}}

{{{cos(A)=1.9999986}}}


{{{A=1.3169571*i}}}(result in radians)

{{{A=75.46*i}}} ° or {{{A=0.0716197}}}° which is very small angle


I was trying to find angle {{{B}}} using side {{{b=9.978}}}, side {{{c=10}}}, and angle {{{C=30}}} using law of sin and got these results:


{{{a = 17.3 }}}  you can get angle {{{C = 30}}}° 
{{{b = 9.98}}}
{{{c = 10}}}

Angles:
{{{A = 120}}}°
{{{B = 29.9}}}°
{{{C = 30}}}° 

so, IF given angle is {{{30}}}, then {{{a = 17.3 }}}

or, if you trying to find angle {{{A}}} using side {{{a=0.025}}}, side {{{c=10}}}, and angle {{{C=30}}, then side {{{b}}} must be {{{10.0216}}}

results are as following:
Sides:
{{{a = 0.025 }}}
{{{b = 10.0216 }}}
{{{c = 10 }}}

Angles:
{{{A = 0.0716197}}}°......I wouldn't prefer this solution because the angle is very, very small 
{{{B = 149.928}}}°
{{{C = 30}}}° 

when you got right sides length, would be possible to find real solutions for angles