Answer
$T_1 = 6396~N$
$T_2 = 4375~N$
Work Step by Step
The magnitude of the horizontal component of the tension in each rope must be equal. Therefore:
$T_1~sin(20^{\circ})=T_2~sin(30^{\circ})$
$T_1=\frac{T_2~sin(30^{\circ})}{sin(20^{\circ})}$
The sum of the vertical components of the tension in each rope must be equal in magnitude to the weight of the steel beam.
$T_1~cos(20^{\circ})+T_2~cos(30^{\circ}) = mg$
$T_2~sin(30^{\circ})~cot(20^{\circ})+T_2~cos(30^{\circ}) = mg$
$T_2 = \frac{mg}{sin(30^{\circ})~cot(20^{\circ})~+~cos(30^{\circ})}$
$T_2 = \frac{(1000~kg)(9.80~m/s^2)}{sin(30^{\circ})~cot(20^{\circ})~+~cos(30^{\circ})}$
$T_2 = 4375~N$
We can use $T_2$ to find $T_1$;
$T_1=\frac{T_2~sin(30^{\circ})}{sin(20^{\circ})}$
$T_1=\frac{(4375~N)~sin(30^{\circ})}{sin(20^{\circ})}$
$T_1 = 6396~N$
