Question

Express the angles in degree, minute and second
(a) ${11.0133}^{\circ }$
(b) ${94.3366}^{\circ }$

Answer 1

We have,

$Degree\times 60=Minutes$

$Minutes\times 60=Seconds$

(a) ${11.0133}^{\circ }$

$=(11+0.0133{)}^{\circ }$

$={11}^{\circ }+(0.0133\times 60{)}^{\prime }$

$={11}^{\circ }+(0.798{)}^{\prime }$

$={11}^{\circ }+(0+0.798{)}^{\prime }$

$={11}^{\circ }+(0{)}^{\prime }+(0.798\times 60{)}^{\prime \prime }$

$={11}^{\circ }+0^{\prime }+{47.88}^{\prime \prime }$

$\therefore {11.0133}^{\circ }={11}^{\circ }{0}^{\prime }{47.88}^{\prime \prime }$ Or ${11}^{\circ }{0}^{\prime }{48}^{\prime \prime }$ approx

(b) ${94.3366}^{\circ }$

$=(94+0.3366{)}^{\circ }$

$={94}^{\circ }+(0.3366\times 60{)}^{\prime }$

$={94}^{\circ }+(20.196{)}^{\prime }$

$={94}^{\circ }+(20+0.196{)}^{\prime }$

$={94}^{\circ }+(20{)}^{\prime }+(0.196\times 60{)}^{\prime \prime }$

$={94}^{\circ }+{20}^{\prime }+{11.76}^{\prime \prime }$

$\therefore {94.3366}^{\circ }={94}^{\circ }{20}^{\prime }{11.76}^{\prime \prime }$ Or ${94}^{\circ }{20}^{\prime }{12}^{\prime \prime }$ approx