alevel mechanics moments help

Hi, can someone help me answer this question please.
A heavy uniform rod AB of mass 25 kg and length 2.4 m is hinged at A to a point on a vertical wall. It is kept horizontal by a chain attached to B and to a point 1.5 m vertically above A. The bar carries an additional mass of 10 kg, 1.8 m from A. The angle that the reaction at A makes with the horizontal is ……
I already found the tension in the chain to be 370N and the magnitude of the reaction at A to be 346N

It would help to see your working. I agree with the tension and reaction magnitude, and in working out the latter you must have resolved horizontally and vertically? If so, the angle should be straightforward?

Resolving vertically i got 147N and horizontally i got 313.7N. Then i did inverse tan 313.7/147 ?

They want the angle with the horizontal. A simple sketch of the force right triangle should get you there.

Im so confused i drew the triangle but now i dont understand which angle im trying to find

If youve worked out all the rest, this part should be straightforward? You have something like
https://homework.study.com/cimages/multimages/16/6799998_-_34581020932779234199.png
At A, the wall exerts a horizontal force of about 300 N on the rod to the right and a vertical force of about 100 N upwards. Adding the two vectors tip to tail gives the usual right triangle which has the overall reaction force on the hypotenuse and the angle it makes with the horizontal is just simple trig.
In the previous post you seemed to work out the angle it makes with the vertical?

ohh i think i get it now, is it just inverse tan 147/313.7 ?