Answer
(a) $11.6\ in$
(b) $37^\circ05'$
Work Step by Step
We know that the raised height should be equal to the arc length traveled by the wheel.
(a) For the wheel, we have the arc length traveled as $L=9.27\times71^\circ50'\times\frac{\pi}{180^\circ}\approx11.6\ in$
(b) For a travel of $6\ in$, we have $6=9.27\times\theta^\circ\times\frac{\pi}{180^\circ}$ which gives $\theta\approx37.085^\circ\approx37^\circ05'$
