$$=(28N)\cos30\degree$$
$$=(28N)(0.866)$$
$$T_x=24.208N$$
$$T_y=Tsin\theta$$
$$=(28N)\sin30\degree$$
$$=(28N)(0.5)$$
$$Ty=14N$$
The magnitude of tension in the leash is given
$$T=\sqrt{T^2_x+T^2_y}$$
$$=\sqrt{(24.208N)^2+(14N)^2}$$
$$T=\sqrt{581.6004N^2+196N^2}$$
$$T=\sqrt{777.6004N^2}$$
$$T=27.89N$$
$$\overrightarrow T=24.208\hat{i}+14\hat{j}$$