$\theta = \arcsin \dfrac{y}{15}$

$\dfrac{d\theta}{dt} = \dfrac{\dfrac{dy/dt}{15}}{\sqrt{1 - \left( \dfrac{y}{15} \right)^2}}$

$\dfrac{d\theta}{dt} = \dfrac{\dfrac{dy/dt}{15}}{\sqrt{1 - \left( \dfrac{y}{15} \right)^2}}$

$\dfrac{d\theta}{dt} = \dfrac{dy/dt}{15\sqrt{1 - \dfrac{y^2}{225}}}$

when $x = 12$

$y = \sqrt{15^2 - 12^2}$

$y = 9 \,\, \text{ ft}$

$\dfrac{d\theta}{dt} = \dfrac{-2}{15\sqrt{1 - \dfrac{9^2}{225}}}$

$\dfrac{d\theta}{dt} = \dfrac{-2}{15\sqrt{1 - \dfrac{81}{225}}}$

$\dfrac{d\theta}{dt} = \dfrac{-2}{15\sqrt{1 - \dfrac{9}{25}}}$

$\dfrac{d\theta}{dt} = \dfrac{-2}{15\sqrt{\dfrac{16}{25}}}$

$\dfrac{d\theta}{dt} = \dfrac{-2}{15 \left( \dfrac{4}{5} \right)}$

$\dfrac{d\theta}{dt} = \dfrac{-2}{12}$

$\dfrac{d\theta}{dt} = -\dfrac{1}{6} \, \text{ rad/sec}$ *answer*

*The negative sign indicates that the angle is decreasing.*