A) \[{{14}^{o}}\]
B) \[{{14}^{o}}19'5''\]
C) \[{{15}^{o}}19'5''\]
D) \[{{16}^{o}}20'\]
Correct Answer: B
Solution :
\[{{\left( \frac{1}{4} \right)}^{C}}=\] \[\left( \frac{1}{4}\times \frac{{{180}^{o}}}{\pi } \right)={{\left( \frac{1}{4}\times \frac{{{180}^{o}}}{22}\times 7 \right)}^{o}}\] \[={{\left( \frac{315}{22} \right)}^{o}}={{\left( 14\frac{7}{22} \right)}^{o}}\] \[={{14}^{o}}{{\left( \frac{7}{22}\times 60 \right)}^{'}}={{14}^{o}}{{\left( 19\frac{1}{11} \right)}^{'}}\] \[={{14}^{o}}19'\left( \frac{1}{11}\times 60 \right)''={{14}^{o}}19'5''\]
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