Tension produced from a string junction When you have three light strings connected at a junction, and you create tension in one of them, as seen here,

(There're masses attached to the ends of the other two cords, not shown in the figure.)
You're supposed to find the tensions produced in the other strings. Using Lami's theorem to find the tension in the other two strings works, as the junction is mass-less and the acceleration produced in it should consequently be zero.
Is there an intuitive way to understand this? Because taking the cosines of the tension in the first string to find this tension doesn't work, and it feels like it should have.
 A: 
Derivation of the Lami's theorem
sum of the forces towards x and y: 
$$\sum_{F_x}=-T_1+T_2\,\cos(\beta')+T_3\,\sin(\gamma')=0\tag 1$$
$$\sum_{F_y}=T_2\,\sin(\beta')-T_3\,\cos(\gamma')=0\tag 2$$
with :
$$\beta'=\pi-\alpha$$
$$\gamma'=\gamma-\frac{\pi}{2}$$
solving eqautions (1) and (2) for $T_1$ and $T_3$ you get:
$$T_3=T_2\frac{\sin(\alpha)}{\sin(\gamma)}\tag 3$$ and
$$T_1=T_2\frac{\sin(\alpha+\gamma)}{\sin(\gamma)}\tag 4$$
equation (3) is the result that we are looking for, 
with:
$$\alpha+\beta+\gamma=2\pi\tag 5$$
solving equation (5) for $\alpha$
$$\alpha=2\pi-\beta-\gamma$$ 
and with equation (4)
$$T_1=T_2\frac{\sin(\beta)}{\sin(\gamma)}\tag 6$$
thus:
with equation (3) and (6) you get the Lami's theorem:
$$\boxed{\frac{T_1}{\sin(\beta)}=\frac{T_2}{\sin(\gamma)}=\frac{T_3}{\sin(\alpha)}}\quad \surd$$
A: Taking components works just fine. For example, let's label the tensions $T_1$, $T_2$, and $T_3$ starting from the horizontal string and going clockwise, and let's align the $x$-axis along the horizontal string. Then the horizontal and vertical components of force balance are
$$T_1 + T_2 \cos \alpha + T_3 \cos \gamma = 0, \quad T_2 \sin \alpha = T_3 \sin \gamma.$$
Solving for $T_2$ in the second equation and plugging it into the first gives
$$T_1 + T_3 \left(\cos \alpha \frac{\sin \gamma}{\sin \alpha} + \cos \gamma \right) = 0$$
so upon simplifying and clearing denominators,
$$T_1 = - T_3 \frac{\cos \alpha \sin \gamma + \sin \alpha \cos \gamma}{\sin \alpha} = - T_3 \frac{\sin(\alpha + \gamma)}{\sin \alpha} = T_3 \frac{\sin \beta}{\sin \alpha}$$
which implies that
$$\frac{T_1}{\sin \beta} = \frac{T_3}{\sin \alpha}.$$
Similarly, by solving for $T_3$ in the second equation and plugging it into the first, we get
$$\frac{T_1}{\sin \beta} = \frac{T_2}{\sin \gamma}.$$
This is precisely Lami's theorem, so the two approaches are equivalent.
