Answer
$250$ mi/h
Work Step by Step
Reference to the right-angled triangle ABC in the fig
$AC=x$
$BC=y$
$\theta=30^\circ$
$c=\sqrt {x^2+y^2}$ ...................... eq (1)
$\frac{dc}{dt}=500$mi/h
$\frac{dx}{dt}=?$
$\tan\theta=\frac{x}{y}$
At $\theta=30^\circ$
$\tan30^\circ=\frac{x}{y}$
Or
$\frac{\sqrt 3}{3}=\frac{x}{y}$
$y=\frac{3x }{\sqrt 3} $ ............... ( 2)
Putting equation (2) in equation (1)
$c=\sqrt {x^2+(\frac{3x }{\sqrt 3})^2}=2x$
Taking derivative with respect to t
$\frac{dc}{dt}=\frac{d(2x)}{dt}=2\frac{dx}{dt}$ ................... (3)
Putting $\frac{dc}{dt}=500$mi/h in equation (3)
$500=2 \frac{dx}{dt}$ Or
$\frac{dx}{dt}=250$ mi/h
