$A,B,C$ satisfy $\sin 2A: \sin 2B: \sin 2C= 5:12:13$ find $A$?

Question

I would appreciate if somebody could help me with the following problem:

Question: $A,B,C$ satisfy (1), (2)

(1). $A+B+C=\pi(0< A,B,C< \pi)$

(2). $\sin 2A: \sin 2B: \sin 2C= 5:12:13$

Find $A$ ?

I tried by using propertyof triangle function formula but couldn’t get it that way.

Answer 1

let $$A'=\pi-2A,B'=\pi-2B,C'=\pi-2C\Longrightarrow A',B',C\in (-\pi,\pi)$$,so $$A'+B'+C'=\pi\Longrightarrow \max{(A',B',C')}>0\tag{1}$$ so in $\Delta A'B'C'$ we have $$\sin{A'}:\sin{B'}:\sin{C'}=\sin{(2A)}:\sin{(2B)}:\sin{(2C)}=5:12:13\tag{2}$$ combine $(1),(2)$ we have $A',B',C'\in (0,\pi)$ and Note $$5^2+12^2=13^2$$then in $\Delta A'B'C'$,we have $$\angle C'=\dfrac{\pi}{2}$$ so $$\sin{A'}=\dfrac{5}{13}=\sin{(2A)}$$ so $$A=\dfrac{1}{2}\arcsin{\dfrac{5}{13}}$$

Answer 2

So, we have $$\sin^22A+\sin^22B=\sin^22C$$

$$\iff\sin^22A=\sin^22C-\sin^22B=\sin(2C+2B)\sin(2C-2B)$$

$$2C+2B=2\pi-2A\implies\sin(2C+2B)=-\sin2A$$

$$\implies\sin2A[-\sin(2C+2B)]=-\sin2A\sin(2C-2B)$$

$$\implies\sin2A[\sin(2C+2B)-\sin(2C-2B)]=0$$

$$\implies\sin2A[2\sin2B\cos2C]=0$$

But $\sin2A\sin2B\ne0$

So, we shall get the value of $C$ directly.

Then use the given ratio to find $\sin2A,$ then $A$