Disclaimer. This is a partial answer, namely for the case that $\,b > 0$ .
All quantities $\,a,b,c\,$ in the problem are integers,
the range of the sine is restricted to $\,0 < \sin < 1$
and the reference value is $\;1+4\sin(10^\circ) \approx 1.7\;$ .
This makes that all of the available choices of the variables maybe aren't that
numerous . It's easy to see that $\,a\,$ cannot be anything else than $a = 1$
for $b > 0$ , if we deliberately ignore the case $\;b < 0\;$ for the moment being. The minimal value of the sine
is $\,\sin(1^\circ)$ , so the possible values of $b > 0$ are bounded by $\;4\sin(10^\circ)/\sin(1^\circ) < 40$ . The
only thing left is to loop through all possible values of $\;c^\circ\;$
and see what happens. The Pascal program below is our implementation of the ideas.
program math110;
procedure medium_force;
const
eps : double = 1.E-6;
var
b,c,max,tel : integer;
value,ref,min : double;
begin
ref := 1+4*sin(10*Pi/180);
Writeln('1+4.sin(10) = ',ref);
min := sin(1*Pi/180);
max := Trunc(4*sin(10*Pi/180)/min)+1;
Writeln('0 < b < ',max);
Writeln;
tel := 0;
for b := 1 to max do
begin
for c := 1 to 89 do
begin
tel := tel + 1;
value := 1+b\*sin(c\*Pi/180);
if (abs(value - ref) < eps) and not (value = ref)
then Writeln('value <> reference'); { Safety check }
if value = ref then Writeln('value = reference');
if value = ref then Writeln('for a = ',1,' b = ',b,' c = ',c);
if value > ref then Break; { Not a chance anymore }
end;
end;
Writeln;
Writeln('# trials = ',tel);
end;
begin
medium_force;
end.
Output:
1+4.sin(10) = 1.69459271066772E+0000
0 < b < 40
value = reference
for a = 1 b = 4 c = 10
\# trials = 196
Note that a "safety check" has been built in against machine rounding errors.
Herewith it is proved that $\;a=1,b= 4,c=10\;$ is indeed a unique solution,
provided that $\,b > 0$ . Now heading for the case $\,b < 0$ , eventually.