#### Answer

$T_x = 281~N$
$T_y = 351~N$

#### Work Step by Step

We can find the angle $\theta$ that the chains make with the vertical.
$cos(\theta) = \frac{2.5~m-0.55~m}{2.5~m}$
$\theta = arccos(\frac{2.5~m-0.55~m}{2.5~m})$
$\theta = 38.7^{\circ}$
We can find the horizontal component of the tension.
$T_x = T~sin(\theta)$
$T_x = (450~N)~sin(38.7^{\circ})$
$T_x = 281~N$
We can find the vertical component of the tension.
$T_y = T~cos(\theta)$
$T_y = (450~N)~cos(38.7^{\circ})$
$T_y = 351~N$