Question

Find the supplementary angle of ${75}^{\circ }{23}^{\prime }{42}^{\prime \prime }$

Answer 1

We write ${180}^{\circ }={179}^{\circ }{59}^{\prime }{60}^{\prime \prime }$

Supplement of ${75}^{\circ }{23}^{\prime }{42}^{\prime \prime }$

$={180}^{\circ }-{75}^{\circ }{23}^{\prime }{42}^{\prime \prime }$

$={179}^{\circ }{59}^{\prime }{60}^{\prime \prime }-{75}^{\circ }{23}^{\prime }{42}^{\prime \prime }$

$={104}^{\circ }{36}^{\prime }{18}^{\prime \prime }$