Would someone be kind enough to check this for me?

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The discussion revolves around measuring string tensions in a kite-flying contest using a weight hanger and protractor. The method involves creating equilibrium by adding weights until the string becomes horizontal, allowing for the calculation of tension based on the mass and angle of the string. The user successfully calculated the tension but is uncertain about the angle used in their calculations, suspecting they may have misinterpreted it. They seek clarification on both the angle's application and how to effectively explain the method to non-physics parents. Overall, the user demonstrates a solid understanding of the physics involved but struggles with communication and angle interpretation.
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Here is the problem (It's a lot of explanation):

You are a judge in a children's kite-flying contest, and two children will win prizes for the kites that pull most strongly and least strongly on their strings. To measure string tensions, you borrow a weight hanger, some slotted weights, and a protractor from your physics teacher, and use the following protocol, illustrated in P4.20. Wait for a child to get her kite well controlled, hook the hanger onto the kite string about 30 cm from her hand, pile on weight until that section of string is horizontal, record the mass required, and record the angle between the horizontal and the string running up to the kite.

a) Explain how this method works. As you construct your explanation, imagine that the children's parents ask you about your method, that they might make false assumptions about your ability without concrete evidence, and that your explanation is an opportunity to give them confidence in your evaluation technique.

b) Find the string tension if the mass is 132 g and the angle of the kite string is 46.3 degrees.

Here is what I did:

MY WORK FOR PART B
I set up my free body diagram and find the x and y components of the string that is going up to the kite (I'm calling this tension T3, the tension of the weights as T1, and the horizontal string T2):

(Fnet)y=ma (a=0 cause its not moving, I'm assuming)
So, T3sin133.7-1.29 N = 0
T3=1.29N/sin133.7 = 1.78 N

Then I plugged T3 into my equation for (Fnet)x:

(Fnet)x=T3cos133.7-T2 = 0
T3cos133.7=T2
1.78Ncos133.7 = T2
-.1.23N = T2So, now I have the tensions of all 3 strings. The only part I'm not really sure about is the angle. Since it says the angle of 46.3 is between the horizontal string and the string going up to the kite, I assumed in order to find the x and y components of T2 I would have to do 180-46.3=133.7. However, for some reason, this seems like a really big kite angle to me and I want to say I'm doing it wrong.

I would love it if someone could tell me if I'm write or wrong about the angle, because I'm pretty sure I did the rest of the work correctly.

MY WORK FOR PART A
Unfortunately, I can't figure out a way to explain this. Obviously, I can grasp the concept since I was able to solve the problem with few worries. However, I can't find a way to write out exactly what I'm doing in sentences so it would make sense to the layperson.

The way I would like to explain it:

I added the weights to the strings in order to create an equilibrium force so that the place where the weights were added has no acceleration and then I used the mass of the weights to figure out the force it exerted in N (or the tension). Once, I found this force I was able to break up the string going up to the kite into x and y components to figure out the tension of of that string. Once I had figured out that tension, I was able to plug that into a physics equation to solve for the tension of the horizontal string.

However, this kind of sounds like crap to me and I don't know if it really would explain the process to someone who isn't in a physics class. Can anyone help me out with this?
 
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for a) find relationship between theta and the tension in the string. you can explain this by givin example of rope used to hang clothes, like how the angle is related to the load of the clothes...

and for b) I also got same thing.
 
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