Question

What is the supplementary angle of ${120}^{\circ }{25}^{\prime }{75}^{\prime \prime }$?

• A. ${59}^{\circ }{45}^{\prime }{33}^{\prime \prime }$
• B. ${60}^{\circ }{33}^{\prime }{45}^{\prime \prime }$
• C. ${59}^{\circ }{33}^{\prime }{45}^{\prime \prime }$
• D. ${60}^{\circ }{45}^{\prime }{33}^{\prime \prime }$

Answer 1

The correct option is C

${59}^{\circ }{33}^{\prime }{45}^{\prime \prime }$

Supplementary angle of $\left({120}^{\circ }{25}^{\prime }{75}^{\prime \prime }\right)={180}^{\circ }-\left({120}^{\circ }{25}^{\prime }{75}^{\prime \prime }\right)$

We can’t subtract 75” from 00”. So, we take $1^{\circ }$ carry from ${180}^{\circ }$. $1^{\circ }={60}^{\prime }$, from which 1’ can be further divided into 60”. But we still can’t subtract 75’’ from 60’’. So, we will take one more carry from remaining 59’. So, we take ${180}^{\circ }={179}^{\circ }{58}^{\prime }120”$ and then proceed with subtraction.

$\begin{array}{c}{179}^{\circ }{58}^{{}^{\prime }}{120}^{\prime \prime }\\ -\underset{–}{{120}^{\circ }{25}^{{}^{\prime }}{75}^{\prime \prime }}\\ \underset{–}{{59}^{\circ }{33}^{{}^{\prime }}{45}^{\prime \prime }}\end{array}$

Hence, complementary angle of $\left({120}^{\circ }{25}^{\prime }{75}^{\prime \prime }\right)={59}^{\circ }{33}^{\prime }{45}^{\prime \prime }$