A year ago, I posted this problem on MSE. Now I have the following similar problem.

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

Question 2:

Assmue that $A,B,C$ be give postive integer numbers,and such $\sin{(\pi\cdot C)}\notin Q$,and $$A+B\sin{(\pi\cdot C )}=a+b\sin{(\pi \cdot c)},a>0,a,b,c\in Z$$ Find $a,b,c$?

I think this problem is not easy (even though it seems clear), maybe we can solve this problem using algebraic number theory? (However, I am not familiar with this field.) If we can prove that $c=10$ is a solution, then I can solve this problem, because if not $a=1$, $b=4$, then $$1+4\sin{10^{0}}=a+b\sin{10^{0}}\Longrightarrow \sin{10^{0}}=\dfrac{a-1}{4-b}\in \mathbb{Q},$$ a contradiction.

Indeed, there is an elementary proof that $\sin{10^{0}}$ is irrational number, or else $$\sin{10^{0}}=\dfrac{q}{p},(p,q)=1,p,q\in \mathbb{N}^{+}.$$ 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 have $$q^2~|~8p^3,~~~\textbf{since}~~~,\gcd{(p,q)}=1,\Longrightarrow q^2|8\Longrightarrow q\le 2$$ so $q=2$. Then we have $p^3-3p+1=0$ and it is clear this equation have no postive integer solution.

Now how to prove that $c=10$ is unique?