1. **State the problem:** We have a right triangle with hypotenuse 13, base $x$, and angle $\theta$ opposite the base. Given $\frac{dx}{dt} = 2$ units/sec, find $\frac{d\theta}{dt}$ when $x=12$. 2. **Formula and relationship:** Using trigonometry, $\sin(\theta) = \frac{x}{13}$. 3. **Differentiate both sides with respect to time $t$:** $$\frac{d}{dt}[\sin(\theta)] = \frac{d}{dt}\left[\frac{x}{13}\right]$$ 4. Applying chain rule and constant multiple rule: $$\cos(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{13} \cdot \frac{dx}{dt}$$ 5. Solve for $\frac{d\theta}{dt}$: $$\frac{d\theta}{dt} = \frac{1}{13} \cdot \frac{\frac{dx}{dt}}{\cos(\theta)}$$ 6. Find $\cos(\theta)$ when $x=12$ using Pythagoras: $$\text{Opposite side} = x = 12$$ $$\text{Hypotenuse} = 13$$ $$\text{Adjacent side} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5$$ 7. Since $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{13}$, substitute values: $$\frac{d\theta}{dt} = \frac{1}{13} \cdot \frac{2}{\frac{5}{13}} = \frac{1}{13} \cdot 2 \cdot \frac{13}{5}$$ 8. Simplify: $$\frac{d\theta}{dt} = \frac{2}{\cancel{13}} \cdot \frac{\cancel{13}}{5} = \frac{2}{5} = 0.4$$ **Final answer:** $$\frac{d\theta}{dt} = 0.4 \text{ radians per second}$$