Answer
$-\frac{1}{8}$
Work Step by Step
Using the chain rule it follows that:
$\frac{d \theta}{dt}=\frac{d \theta}{dx}\cdot \frac{dx}{dt}$
$\frac{d \theta}{dt}|_{x=20}=\frac{d \theta}{dx}|_{x=20}\cdot 5$
Find $\frac{d \theta}{dx}|_{x=20}$.
Since the triangle is right it follows that:
$$\tan(\theta)=\frac{20}{x}$$
Differentiate both sides it follows:
$$(\tan(\theta))'=\left(\frac{20}{x}\right)'$$
Use the chain rule it follows:
$$\frac{d\theta}{dx}\cdot \frac{1}{\cos^{2}(\theta)}=\left(\frac{20}{x}\right)'$$
$$\frac{d\theta}{dx}\cdot \frac{1}{\cos^{2}(\theta)}=\frac{-20}{x^{2}}$$
$$\frac{d\theta}{dx}=\frac{-20}{x^{2}}\cdot \cos^{2}(\theta)$$
Since the triangle is right it follows that:
$$\cos(\theta)=\frac{x}{\sqrt{x^{2}+20^{2}}}$$
so:
$$\frac{d\theta}{dx}=\frac{-20}{x^{2}}\cdot \left(\frac{x}{\sqrt{x^{2}+20^{2}}}\right)^{2}$$
$$\frac{d\theta}{dx}=\frac{-20}{x^{2}}\cdot \frac{x^{2}}{x^{2}+20^{2}}$$
$$\frac{d\theta}{dx}=-20\cdot \frac{1}{x^{2}+20^{2}}$$
$$\frac{d\theta}{dx}|_{x=20}=-20\cdot \frac{1}{20^{2}+20^{2}}=-\frac{1}{40}$$
so:
$\frac{d \theta}{dt}|_{x=20}=-\frac{1}{40}\cdot 5=-\frac{1}{8}$
