Find the degree measure corresponding to the following radian measures:

Solution:

We have :

$\pi \operatorname{rad}=180^{\circ}$

$\therefore 1 \mathrm{rad}=\left(\frac{180}{\pi}\right)^{\circ}$

(i) $\frac{18 \pi}{5}=\left(\frac{180}{\pi} \times \frac{9 \pi}{5}\right)^{\circ}$

$=(36 \times 9)^{\circ}$

$=324^{\circ}$

(ii) $-\frac{5 \pi}{6}=\left(\frac{180}{\pi} \times\left(-\frac{5 \pi}{6}\right)\right)^{\circ}$

$=-(30 \times 5)^{\circ}$

$=-(150)^{\circ}$

(iii) $\left(\frac{18 \pi}{5}\right)^{c}=\left(\frac{180}{\pi} \times \frac{18 \pi}{5}\right)^{\circ}$

$=(36 \times 18)^{\circ}$

$=648^{\circ}$

(iv) $(-3)^{c}=\left(\frac{180}{\pi} \times-3\right)^{\circ}$

$=\left(\frac{180}{22} \times 7 \times-3\right)^{\circ}$

$=\left(\frac{-3780}{22}\right)^{\circ}$

$=\left(-171 \frac{18}{22}\right)^{\circ}$

$=\left\{-171^{\circ}\left(\frac{18}{22} \times 60\right)^{\prime}\right\}$

$=\left\{-171^{\circ}\left(49 \frac{1}{11}\right)^{\prime}\right\}$

$=-\left\{171^{\circ} 49^{\prime}\left(\frac{1}{11} \times 60\right)^{\prime \prime}\right\}$

$=-\left(171^{\circ} 49^{\prime} 5.45^{\prime \prime}\right)$

$\approx-\left(171^{\circ} 49^{\prime} 5^{\prime \prime}\right)$

(v) $(11)^{c}=\left(\frac{180}{\pi} \times 11\right)^{\circ}$

$=\left(\frac{180}{22} \times 7 \times 11\right)^{\circ}$

$=630^{\circ}$

(vi) $(1)^{c}=\left(\frac{180}{\pi} \times 1\right)^{\circ}$

$=\left(\frac{180}{22} \times 7 \times 1\right)^{\circ}$

$=\left(\frac{630}{11}\right)^{\circ}$

$=\left(57 \frac{3}{11}\right)^{\circ}$

$=57^{\circ}\left(\frac{3}{11} \times 60\right)^{\prime}$

$=57^{\circ}\left(16 \frac{4}{11}\right)^{\prime}$

$=57^{\circ} 16^{\prime}\left(\frac{4}{11} \times 60\right)^{\prime \prime}$

$=57^{\circ} 16^{\prime} 21.81^{\prime \prime}$

$\approx 57^{\circ} 16^{\prime} 22^{\prime \prime}$