Question

Add:
${56}^{\circ }{38}^{\prime }$ and ${27}^{\circ }{42}^{\prime }{30}^{\prime \prime }$

Answer 1

Adding the given angles

Addition of ${56}^{\circ }{38}^{\prime }$ and ${27}^{\circ }{42}^{\prime }{30}^{\prime \prime }$

${56}^{\circ }{38}^{\prime }{00}^{\prime \prime }$
$\underset{–}{+{27}^{\circ }{42}^{\prime }{30}^{\prime \prime }}$
$={83}^{\circ }{80}^{\prime }{30}^{\prime \prime }$

Now, we know that ${60}^{\prime }=1^{\circ }$
$\therefore {80}^{\prime }={60}^{\prime }+{20}^{\prime }=1^{\circ }{20}^{\prime }$
$\Rightarrow {83}^{\circ }{80}^{\prime }{30}^{\prime \prime }={84}^{\circ }{20}^{\prime }{30}^{\prime \prime }$

Hence, ${84}^{\circ }{20}^{\prime }{30}^{\prime \prime }$ is the required result of addition.