Question

Express in Degrees:
$\left(a\right){\left(\frac{2\pi }{15}\right)}^c$

$\left(b\right)(-2{)}^c$

• A. $\left(a\right){14}^{\circ }$
  $\left(b\right){244}^{\circ }{32}^{\prime }{44}^{\prime \prime }$
• B. $\left(a\right){74}^{\circ }$
  $\left(b\right)-{114}^{\circ }{32}^{\prime }{44}^{\prime \prime }$
• C. $\left(a\right){14}^{\circ }$
  $\left(b\right)-{120}^{\circ }{32}^{\prime }{44}^{\prime \prime }$
• D. $\left(a\right){24}^{\circ }$
  $\left(b\right)-{114}^{\circ }{32}^{\prime }{44}^{\prime \prime }$

Answer 1

The correct option is D $\left(a\right){24}^{\circ }$

$\left(b\right)-{114}^{\circ }{32}^{\prime }{44}^{\prime \prime }$
(a) ${\left(\frac{2\pi }{15}\right)}^c={(\frac{2\pi }{15}\times \frac{180}{\pi })}^{\circ }={24}^{\circ }$
(b) $(-2{)}^c={(-2\times \frac{180}{\pi })}^{\circ }$
$={(\frac{180}{22}\times 7\times (-2\left)\right)}^{\circ }={(-114\frac{6}{11})}^{\circ }$
$=-114{(\frac{6}{11}\times 60)}^{\circ }=-{114}^{\circ }\left(32\frac{8}{11}\right)$
$=-{114}^{\circ }{32}^{\prime }(\frac{8}{11}\times 60)$
$=-{114}^{\circ }{32}^{\prime }{44}^{\prime \prime }$