The question given is
A uniform rod, of mass $m$ and length $2a$ smoothly hinged to a vertical wall is connected to a point on the wall above the hinge by a light inelastic string. Find the magnitude and direction of the force on the rod from the hinge.
The diagram given is
I have constructed the forces on to the freebody diagram as such
Using the vertical component of net force: $$F_x=T \sin(20)-mg+R_V=0$$ $$\implies T\sin(20)=mg-R_V$$
Using the horizontal component of net force: $$F_y=T \cos(20)-R_H=0$$ $$\implies T\cos(20)=R_H$$ $$\implies T=\frac{R_H}{\cos(20)}$$
Finding the moment around the hinge A: $$\tau=T\sin(20)*2a\sin(50)-mg*a\sin(50)=0$$ $$\implies T\sin(20)=\frac{mg}{2}$$ $$\implies T=\frac{mg}{2\sin(20)}$$
Then, equating the two equations for $T\sin(20)$ I get: $$\frac{mg}{2}=mg-R_V$$ $$\implies 2R_V=mg$$
Finally equating both equations for $T$: $$\frac{mg}{2\sin(20)}=\frac{R_H}{\cos(20)}$$ $$\implies \frac{2R_V}{2\sin(20)}=\frac{R_H}{\cos(20)}$$ $$\implies \frac{R_V}{R_H}=\frac{\sin(20)}{\cos(20)}$$
And since $R_V,R_H$ are the vertical and horizontal components of the same force respectively, the $tan^{-1}$ of this ratio is equal to the angle between the resultant and the horizontal hence:
$$\tan^{-1}(\frac{\sin(20)}{\cos(20)})=20°$$
Which is wrong as the answer given in the book is that the angle is $82.7°$ and the reaction force is $0.474mg$, I am aware that this entire method is a hackjob but I have no clue on how to tackle this question properly, are the reaction forces I drew the problem or is it my working onwards?
How would I tackle this problem properly?
