Answer:
[tex]\dot \theta = 0.144\,\frac{rad}{s}[/tex], [tex]\dot \theta = 8.251\,\frac{deg}{s}[/tex] (Option B)
Step-by-step explanation:
The trigonometric diagram is included herein as attachment. The expression is presented below:
[tex]\tan \theta = \frac{y}{x}[/tex]
Where:
[tex]x[/tex] - Horizontal distance between the geyser and the camera.
[tex]y[/tex] - Vertical distance between the geyser and the camera.
The rate of change in terms of time is:
[tex]\dot \theta \cdot \sec^{2}\theta = \frac{\dot y\cdot x-y\cdot \dot x}{x^{2}}[/tex]
[tex]\dot \theta \cdot \left(\frac{1}{\cos^{2}\theta} \right) = \frac{\dot y \cdot x - y\cdot \dot x}{x^{2}}[/tex]
[tex]\dot \theta = \left(\frac{\dot y \cdot x - y \cdot \dot x}{x^{2}} \right)\cdot \cos^{2}\theta[/tex]
[tex]\dot \theta = \left(\frac{\dot y \cdot x - y\cdot \dot x}{x^{2}} \right)\cdot \left(\frac{x^{2}}{x^{2}+y^{2}} \right)[/tex]
[tex]\dot \theta = \frac{\dot y \cdot x - y\cdot \dot x}{x^{2}+y^{2}}[/tex]
Finally,
[tex]\dot \theta = \frac{\left(900\,\frac{ft}{s} \right)\cdot (4000\,ft)-(3000\,ft)\cdot \left(0\,\frac{ft}{s} \right)}{(4000\,ft)^{2}+(3000\,ft)^{2}}[/tex]
[tex]\dot \theta = 0.144\,\frac{rad}{s}[/tex]
[tex]\dot \theta = 8.251\,\frac{deg}{s}[/tex]
