First thing first, that is Geometry. We need to know the angle the cable makes with the vertical (and the bridge), in order to find its magnitude and components.

Let’s designate some points: A is where the cable attaches to the bridge, O the hinge of the bridge, B where the cable attaches to the castle wall. We have, in the triangle AOB, OB = 12 m, OA = 5 m.

Now, OB is vertical and OA is 20 deg below horizontal, that makes angle AOB 110 deg, hence, angles OAB + OBA (call them A and B) is supplement to 110 deg:
A + B = 70 or A = 70 – B

Also, applying Sine Rule of Triangles:
sinA / 12 = sinB / 5
12 sinB = 5 sin(70 – B) = 5 sin 70 cosB – 5 cos70 sinB
tanB = …
Solving, B = … deg, A = … deg

Now that Geometry is out of the way, let’s delve into Statics. Moment of a force about a point is the cross (outer) product of the force and radius vector (line from point of moment to point of action of force). If we take moment about the hinge, the moment of the cable tension must balance the moments of the weights of the bridge and the knight and stead.

Also if the angle between the bridge and horizontal is 20 deg, then angle between bridge and vertical (direction of weight) must be 70 deg. So here we go
T x 5 sin 51.1 = 2000 g x 4 sin 70 + 1200 g x 7 sin 70
where T is cable tension
Solving T = … N

Only horizontal force causing a reaction at the hinge is the horizontal component of the tension (the weights are vertical)
H = T sin B = … N

Direction: opposite to horizontal component of T.

Vertical component of tension = T cos A = … N
That is not the only vertical force on the bridge, there are the weights to consider. So
V + T cos B – 2000 g – 1200 g = 0
V = … N

Direction: figure out from sin of V, positive upwards..

A = 51.1, B = 28.9, T = 38852 N, H = 12585 N, compression, V = -5365 N, downwards.
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