Part B
In the given right triangle
we have that
Applying the Pythagorean Theorem
[tex]\begin{gathered} L^2=x^2+6^2 \\ L=\sqrt{x^2+36} \end{gathered}[/tex][tex]\begin{gathered} cos\theta=\frac{6}{L} \\ cos\theta=\frac{6}{\sqrt{x^2+36}} \\ \\ \theta=cos^{-1}(\frac{6}{\sqrt{x^2+36}}) \\ \end{gathered}[/tex]Find out the derivative
[tex]\frac{d\theta}{dt}=\frac{6}{x^2+36}*\frac{dx}{dt}[/tex]Remember that
dx/dt=5/2 ft/sec (part a)
L^2=x^2+36=10^2
substitute given values
[tex]\frac{d\theta}{dt}=\frac{6}{10^2}*\frac{5}{2}=\frac{15}{100}=0.15[/tex]