a years ago, I have post this problem on [MSE], and then I consider following interesting problem:

>Assmue that：
$$1+4\sin{10^\circ}=a+b\sin{c^\circ}$$
where $a>0$,and $a,b,c$ are integers and $0<c<90^\circ$,
>show that $$a=1,b= 4,c=10$$ is unique solution

I think this problem is not easy,maybe can use algebraic number theory? (because I donit have any knowledge about this),if we can prove $c=10$ is solution,then I can prove $a=1,b=4$,because if
$$1+4\sin{10^{0}}=a+b\sin{10^{0}}\Longrightarrow \sin{10^{0}}=\dfrac{a-1}{4-b}\in Q$$
a contradiction.

Indeed,we can use Elementary proof $\sin{10^{0}}$ is irrational number,
or else
$$\sin{10^{0}}=\dfrac{q}{p},(p,q)=1,p,q\in N^{+}$$
note
$$\dfrac{1}{2}=\sin{30^{0}}=3\sin{10^{0}}-4\sin^3{10^{0}}=3\cdot\dfrac{q}{p}-4\left(\dfrac{q}{p}\right)^3\Longrightarrow 8p^3-6pq^2+q^3=0$$
so we $$q^2|8p^3,\textbf{since},\gcd{(p,q)}=1,\Longrightarrow q^2|8\Longrightarrow q\le 2$$
so $q=2$,then we have $p^3-3p+1=0$,it is clear this equation have no postive integer solution.

By Done!


**Now solve this problem key is How prove $c=10$ is unique ?**


[MSE]:http://math.stackexchange.com/questions/359594/find-this-a-b-c-such-that-sqrt9-8-sin-50-circ-ab-sin-c-circ