Answer
$\theta=37.6^o$
Work Step by Step
The free-body diagram is shown below.
Summing the forces in the y-direction and the x-direction:
$$\mu_sG-N=0$$ $$0.65G-N=0 (2)$$
Next, we choose the axis of rotation to be the point where the ladder and the ground meets, so $G$ and $f_s$ produce no torque.
Take the board's length to be $l$:
$$Nl\sin\theta=\frac{Wl\cos\theta}{2}$$ $$N\sin\theta=\frac{W\cos\theta}{2}$$ $$\frac{N}{W}\tan\theta=\frac{1}{2}$$
From (1), we have $W=G$ and from (2), we have $N=0.65G$. Therefore, $$\frac{0.65G}{G}\tan\theta=\frac{1}{2}$$ $$0.65\tan\theta=\frac{1}{2}$$ $$\tan\theta=0.769$$ $$\theta=37.6^o$$
