A years ago, I have post this problem on MSE, and Now 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 am not familiar with this field),if we can prove $c=10$ is solution,then I can solve this problem:$a=1,b=4$,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 ?