An $$L$$-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If $$AB = BC$$, and the angle made by $$AB$$ with downward vertical is $$\theta$$, then:

Let $$AB = BC = L$$, and the mass of each uniform rod segment be $$m$$.

Let the point of suspension $$A$$ be the origin $$(0,0)$$.

Taking the downward vertical as the reference line. 

Coordinates of the center of mass of rod AB ($$G_1$$): $$x_1 = -\frac{L}{2}\sin\theta, \quad z_1 = -\frac{L}{2}\cos\theta$$

Coordinates of the center of mass of rod BC ($$G_2$$): $$x_2 = -L\sin\theta + \frac{L}{2}\cos\theta, \quad z_2 = -L\cos\theta - \frac{L}{2}\sin\theta$$

For rotational equilibrium about the suspension point $$A$$:

$$\tau_A = 0 \implies m g |x_1| + m g x_2 = 0 \implies x_{\text{cm}} = 0$$

$$m\left(-\frac{L}{2}\sin\theta\right) + m\left(-L\sin\theta + \frac{L}{2}\cos\theta\right) = 0$$

$$-\frac{3}{2}L\sin\theta + \frac{1}{2}L\cos\theta = 0 \implies 3\sin\theta = \cos\theta$$

$$\tan\theta = \frac{1}{3}$$