Hanged bird feeder

Let $T_A$ and $T_B$ be the tensions in wire $A$ and $B$ respectively. Then equating the horizontal forces: $T_Acos60^\circ = T_Bcos30^\circ$ $\Rightarrow T_A = \sqrt{3}T_B$ Now equating vertical forces, we have: $T_Asin60^\circ + T_Bsin30^\circ = 65$ $\Rightarrow \frac{\sqrt{3}}{2} T_A + \frac{1}{2}T_B = 65$ $\Rightarrow \frac{3}{2} T_B + \frac{1}{2}T_B = 65$ $\Rightarrow 2 T_B = 65 \Rightarrow T_B=\boxed{32.5N}$

Equating the forces vertically and horizontally, we have two equations that go like so: Let tension at A be called A and the tension at B be called B (for simplicity)

Asin60+Bsin30=65 (forces vertically)

Acos60=Bcos30 (forces horizontally)

Using the second equation: Acos60=Bcos30

A(0.5)=B(sqrt3/2)

Hence A=(sqrt3) B

Substituting that into the first equation, we get: (sqrt3)B(sqrt3/2)+(0.5)B=65

1.5B+0.5B=65

Hence 2B=65 and B=32.5N

I'm not familiar with all these math editor stuff yet, as this is my first solution post, but I hope everyone gets the solution. I've kept it as simple as possible. Hope everyone understands. Cheers!