Question

refer to the figure to the right. (a) how many inches will the weight in the figure rise if the pulley is rotated through an angle of 76° 50? (b) through what angle, to the nearest minute, must the pulley be rotated to raise the weight 6 in.? (a) the weight in the figure will rise 12.1 inches. (do not round until the final answer. then round to the nearest tenth as needed.) (b) in order for the weight to rise 6 inches, the pulley must rotate through ° . (do not round until the final answer. then round to the nearest minute as needed.)

Explanation:

Step1: Convert angle to radians

First, convert $76^{\circ}50'$ to decimal - degrees. $50'=\frac{50}{60}\approx0.833^{\circ}$, so $76^{\circ}50'\approx76.833^{\circ}$. Then convert to radians: $\theta = 76.833\times\frac{\pi}{180}\approx1.341$ radians. The formula for arc - length $s = r\theta$, where $r = 9.01$ in. So $s=9.01\times1.341\approx12.1$ in.

Step2: Solve for the angle in part (b)

We know $s = r\theta$. Given $s = 6$ in and $r = 9.01$ in, then $\theta=\frac{s}{r}=\frac{6}{9.01}\approx0.666$ radians. Convert this to degrees: $\theta_d=\theta\times\frac{180}{\pi}=\frac{0.666\times180}{\pi}\approx38.26^{\circ}$. Convert the decimal part to minutes. $0.26\times60 = 15.6'$. Rounding to the nearest minute, $\theta\approx38^{\circ}16'$.

Answer:

(a) 12.1
(b) $38^{\circ}16'$