Angle rate for a 50-ft ladder sliding on oil

A ladder $50$ ft long leans on a vertical wall. Its foot sits in oil and slides directly away from the wall at a steady $4$ ft per minute, so the top slips down the wall. At the moment the foot is $30$ ft from the wall, how fast (radians per minute) is the angle between the ladder and the ground shrinking?

$1/10$ rad per minute (i.e. $\theta'=-\tfrac1{10}$). With $x=50\cos\theta$, $x'=-50\sin\theta\,\theta'$. At $x=30$ the top is $40$ ft high, so $\sin\theta=\tfrac45$ and $4=-40\,\theta'$, giving $\theta'=-0.1$.