Problem 10.2 #11

Professor Taylor,
This problem has me stumped.
1/2 * 12 = 6
arctan(3/6) = 26.6 <-- should be using radians instead of degrees

1*9.8=2Tsin(26.6) <-- good 



9.8/2sin(26.6)<-- = T

1/2 (9.8)/sin(26.6)=10.943

I might be applying the wrong formula. But even so, I would
appreciate it if you could help me.
Sincerely,

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You had all the pieces and did most of the calculations correctly. It looks like you just got distracted at the end and lost track of your goal--you needed T and not θ.

First of all, use radians instead of degrees--there's a button on your calculator for this.
Second, always draw a picture--insufficient understanding of the geometry is one of the biggest problems that people have in this course and making a picture is the best solution. This one should look like this:

The force of gravity vector is pointing straight down and has magnitude 9.8m/sec^2 x 1kg=9.8N.  The tension on each half of the line is pointing from the tip of the pole to kink in the line. They make an angle of θ with respect to the horizontal, so as you already understand tan(θ)=3/6=1/2, so
sin(θ)/cos(θ)=1/2 so 2 sin(θ)=cos(θ), and cos^2(θ)+sin^2(θ)=1 so 4 sin^2(θ)+sin^2(θ)=1 so
5sin^2(θ)=1 so sin(θ)=1/√5 so cos(θ)=2/√5---and of course yes θ = 26.6 degrees or 0.464 radians of them, but you only need sin(θ) anyway and you already have that.