Calculus 1 related rates involving trigonometry?

The angle just below the left end of the dotted line is pi/3 radians, so the angle just above that is pi - (pi/3) = 2pi/3 radians. So the angle above the right end of the dotted line is pi - (2pi/3) - ϴ = (pi/3) - ϴ radians. So the Sine Rule gives us:

sin((pi/3) - ϴ) / 12 = sin(2pi/3) / x

sin((pi/3) - ϴ) / 12 = sqrt(0.75) / x

x*sin((pi/3) - ϴ) = 12*sqrt(0.75)

x*sin((pi/3) - ϴ) = sqrt(108)

x = sqrt(108) / sin((pi/3) - ϴ)

dx/dϴ = 6*sqrt(3)*tan(ϴ + (pi/6))*sec(ϴ + (pi/6))

dx/dt = -0.5

By the Chain Rule:

dx/dt = dx/dϴ * dϴ/dt

-0.5 = 6*sqrt(3)*tan(ϴ + (pi/6))*sec(ϴ + (pi/6)) * dϴ/dt

dϴ/dt = -0.5 / (6*sqrt(3)*tan(ϴ + (pi/6))*sec(ϴ + (pi/6)))

When ϴ = pi/6 radians:

dϴ/dt = -0.5 / (6*sqrt(3)*tan((pi/6) + (pi/6))*sec((pi/6) + (pi/6)))

dϴ/dt = -1/72 radians per second

Find how x depends on θ.

Law of sines

x / sin (2π/3) = 12 / sin (π- 2π/3-θ)

x / (√3/2) = 12 / sin (π/3-θ)

x = 6√3 / sin (π/3-θ) 

x = 12√3 / (√3 cos θ - sin θ)

dx/dθ = 12 (3 sin θ + √3 cos θ)/(√3 cos θ - sin θ)²

multiply by dt/dt

(dx/dt)(dt/dθ) = 12 (3 sin θ + √3 cos θ)/(√3 cos θ - sin θ)²

plug in dx/dt=1/2

(dt/dθ) = 24 (3 sin θ + √3 cos θ)/(√3 cos θ - sin θ)²

dθ/dt = (√3 cos θ - sin θ)² / (24 (3 sin θ + √3 cos θ))

this can be simplified a little bit

dθ/dt = (√3/36) (cos(θ + π/6) cot(θ + π/6))

Plug in θ=π/6

dθ/dt (θ=π/6) = (√3/36) (cos(π/3) cot(π/3))

= (√3/36) (1/2) (1/√3)

= 1/72