The free-body diagram of the given problem is the following:
In the diagram we have the following forces:

We are asked to determine the tension in the horizontal tension we will add the horizontal forces:

Since there is no movement in this direction this means that the sum of forces must be equal to zero, therefore, we have:

Solving for the horizontal tension we get:

From the following right triangle we can determine the value of the x-coordinate of the tension "T":
We can use the function cosine and we get:

Now we multiply both sides by T:

Now we substitute this value in the sum of forces:

Now we need to determine the value of "T". To do that we will add the vertical forces, we get:

Since there is no vertical movement the forces add up to zero, we get:

Now we use the same right triangle to get the value of the y-component of the tension:

Multiplying both sides by "T":

Now we substitute in the sum of vertical forces:

Now we solve for "T", first by adding "mg" to both sides:

Now we divide both sides by "sin37":

Now we substitute this value in the formula for the horizontal tension:

Now we substitute the values:
![T_H=\frac{(73\operatorname{kg})(9.8(m)/(s^2))}{\sin37}\cos 37]()
Now we solve the operations:

Therefore, the tension in the horizontal section is 949.37N.