Question

Find the complement of: ${23}^{\circ }{19}^{\prime }{32}^{\prime \prime }$.

Answer 1

We know, two angles are said to be complementary, if the sum of their measures is ${90}^{\circ }.$

Let the angle be $x^{\circ }$ and its complement is ${90}^{\circ }-x^{\circ }$

Then, the complement of $x={23}^{\circ }{19}^{\prime }{32}^{\prime \prime }$

$={90}^{\circ }-x^{\circ }$

$={90}^{\circ }-{23}^{\circ }{19}^{\prime }{32}^{\prime \prime }$

$={89}^{\circ }{59}^{\prime }{60}^{\prime \prime }-{23}^{\circ }{19}^{\prime }{32}^{\prime \prime }$

$={66}^{\circ }{40}^{\prime }{28}^{\prime \prime }$.

Hence, the complement is ${66}^{\circ }{40}^{\prime }{28}^{\prime \prime }$.