Surely you know how to **add**! That is what the first is asking you to do- together with know that there are 60 seconds in a minute and 60 minutes in a degree.

To add \(\displaystyle 13^o 45' 30''+ 20^0 25' 40''\)

30"+ 40"= 70"= 60"+ 10"= 1' 10"

45'+ 25'= 70' plus that additional minute from the seconds sum is 71'. 71'= 60'+ 11'= \(\displaystyle 1^o 11'

\(\displaystyle 13^o+ 20^o= 33^o\) plus that additional degree from the minutes sum is \(\displaystyle 34^o\).

The sum is \(\displaystyle 34^o 11' 10"\)

To write \(\displaystyle 30.2^o\) "without a decimal", convert that "\(\displaystyle .2^o\)" to minutes and (if necessary) seconds.

Since there are \(\displaystyle 60'\) in a degree, .2 of a degree is (0.2)(60)= 3' so \(\displaystyle 30.2^o= 30^0 3'\).\)