Answer
$7.1~in/min$
Work Step by Step
Step 1. Using the figure in the Exercise, take the bottom of the building as the origin; we have $x=60ft, y=80ft$ and $\frac{d\theta}{dt}=0.27^{\circ}/min\approx0.0047rad/min$
Step 2. $tan\theta=\frac{80}{x}$ and $sec^2\theta\frac{d\theta}{dt}=-\frac{80}{x^2}\frac{dx}{dt}$
Step 3. We have $cos\theta=\frac{x}{\sqrt {x^2+y^2}}=\frac{60}{\sqrt {60^2+80^2}}=\frac{3}{5}$
Step 4. Equation of step 2 becomes $\frac{25}{9}(0.0047)=-\frac{80}{60^2}\frac{dx}{dt}$ which gives
$\frac{dx}{dt}=-\frac{25(0.0047)(60^2)}{9(80)}\approx0.589ft/min\approx7.1in/min$
