MHB Solving for AB & AC: Find the Solution

[paulmdrdo1]

Find AB and AC in the figure given.

I came up with 2 equations which have 3 unknowns,

$AB\sin(55)+AC\sin(30)=BC$
$AB\sin(35)=AC\sin(60)$

I'm not sure if I'm just missing some given in the problem. Because if BC is given I can solve the equation above. Please help. thanks!

 

[MarkFL]

First, let's resolve the 3 forces into their $x$ and $y$ components (where point $D$ is directly below $A$, the mass):

[table="width: 400, class: grid, align: left"]
[tr]
[td]Force[/td]
[td]$x$-component[/td]
[td]$y$-component[/td]
[/tr]
[tr]
[td]$\vec{AB}$[/td]
[td]$-\vec{AB}\sin\left(55^{\circ}\right)$[/td]
[td]$\vec{AB}\cos\left(55^{\circ}\right)$[/td]
[/tr]
[tr]
[td]$\vec{AC}$[/td]
[td]$\vec{AC}\sin\left(30^{\circ}\right)$[/td]
[td]$\vec{AC}\cos\left(30^{\circ}\right)$[/td]
[/tr]
[tr]
[td]$\vec{AD}$[/td]
[td]$0$[/td]
[td]$-360\text{ N}$[/td]
[/tr]
[/table]

Now, the first condition for equilibrium gives us the equations:

$$\sum F_x=\vec{AC}\sin\left(30^{\circ}\right)-\vec{AB}\sin\left(55^{\circ}\right)=0$$

$$\sum F_y=\vec{AB}\cos\left(55^{\circ}\right)+\vec{AC}\cos\left(30^{\circ}\right)-360\text{ N}=0$$

Now, you have two equations in two unknowns. :D