Answer & Solution
Answer: Option B
Solution:
$$\eqalign{
& {180^ \circ } = \pi \,{\text{radian}} \cr
& 1{\text{ radian}} = \frac{{{{180}^ \circ }}}{\pi } \cr
& \frac{{{{180}^ \circ }}}{\pi } \Rightarrow \frac{{180 \times 7}}{{22}} \Rightarrow \frac{{630}}{{11}} \Rightarrow 57\frac{{{3^ \circ }}}{{11}} \cr
& = \frac{1}{{\cot }} = {57^ \circ }\frac{{180'}}{{11}} \cr
& = {57^ \circ }16'\frac{{4'}}{{11}} \cr
& = {57^ \circ }16'\left( {\frac{4}{{11}} \times 60''} \right) \cr
& = {57^ \circ }16'22'' \cr} $$
