QUESTION (see attached image file):
I got stuck when trying to find tension for part A; see my working below. and help me solve.
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It looks like you're grasped the basic concept of the question -- using trigonometry to solve for the inclination of the cable, and then summing moments about point C to find the tension in the cable.
From the trig, angle alpha is 47.6025 degrees (as shown in sketch below).
Then, summing moments about point C, the tension T is calculated.
The problem is a trigonometric problem. The trick is for you to solve the value of A, B, C, and x.
To solve for x, you'll need the cosine law such that:
Angle C is very obvious to be equal to 70 degrees.
$$x^2 = 10^2 + 12^2 - 2(10)(12)cos(70)$$
$$x = 12.7246 \ cm\ = 12.72 cm.$$
now, solve for the value of angle (A+B). Angle A is equal to 20 degrees.
$$cos (A+B) = (12^2 - 10^2 - 12.72^2) / (-2*10*12.72)$$
therefore, $$(A+B) = 62.416\ degrees$$
Now, rotating the FBD by 20 degrees clockwise will give the image above. Summing up moments along the hinge (C) will give:
$$T sin(62.416) * 10 = 200 sin (70) * 15$$
$$T = 318.06 \ N \ ---->answer$$