Question

$\text{find the sum of}{52}^{\circ }{38}^{\prime }\text{and}{37}^{\circ }{42}^{\prime }{30}^{\prime \prime }.$

• A. ${89}^{\circ }{20}^{\prime }30"$
• B. ${90}^{\circ }{20}^{\prime }30"$
• C. ${90}^{\circ }{10}^{\prime }30"$
• D. ${89}^{\circ }{80}^{\prime }$

Answer 1

The correct option is B ${90}^{\circ }{20}^{\prime }30"$

${52}^{\circ }+{37}^{\circ }=89{38}^{\prime }+{42}^{\prime }={80}^{\prime }00"+{30}^{\prime \prime }={30}^{\prime \prime }$

$\text{since}{38}^{\prime }+{42}^{\prime }={80}^{\prime },{60}^{\prime }=1^{\circ }\text{has to taken over to degrees as carry, as minutes can only vary form 1 to}\text{60. they can't take a value more than 60. Hence 60' or 1°will be carried over and 20' will be left behind. Now}{1}^{\circ }\text{will be added to}{89}^{\circ }\text{which is the sum of}{52}^{\circ }\text{and}{37}^{\circ }.$

$\text{So, sum =}{90}^{\circ }{20}^{\prime }{30}^{\prime \prime }$