Exercise 1

Convert the following units into sexagesimal form:

8,179''

27,520''

The best Maths tutors available
1st lesson free!
Ayush
5
5 (27 reviews)
Ayush
£90
/h
1st lesson free!
Intasar
4.9
4.9 (23 reviews)
Intasar
£42
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (6 reviews)
Dr. Kritaphat
£39
/h
1st lesson free!
Paolo
4.9
4.9 (11 reviews)
Paolo
£25
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Myriam
5
5 (15 reviews)
Myriam
£20
/h
1st lesson free!
Andrea
5
5 (12 reviews)
Andrea
£40
/h
1st lesson free!
Ayush
5
5 (27 reviews)
Ayush
£90
/h
1st lesson free!
Intasar
4.9
4.9 (23 reviews)
Intasar
£42
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (6 reviews)
Dr. Kritaphat
£39
/h
1st lesson free!
Paolo
4.9
4.9 (11 reviews)
Paolo
£25
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Myriam
5
5 (15 reviews)
Myriam
£20
/h
1st lesson free!
Andrea
5
5 (12 reviews)
Andrea
£40
/h
First Lesson Free>

Exercise 2

Convert the following units into decimal form:

12° 30' 42''.

Exercise 3

Express the following angles in degrees:

3 rad

\frac {2 \pi}{5} rad

\frac {\pi}{10} rad

Exercise 4

Express the following angles in radians:

316°

10°

127º

Exercise 5

Calculate:

68º 35' 42'' + 56º 46' 39''.

Exercise 6

Calculate:

 (132° 26' 33'') × 5

 (128° 42' 36'') × 3

 

Exercise 7

Calculate:

(132° 26' 33'') : 3.

Exercise 8

Determine the complementary and supplementary angle of 38° 36' 43''.

Exercise 9

Determine the complementary and supplementary angle of 25° 38' 40''.

 

Solution of exercise 1

Convert the following units into sexagesimal form:

8,179''

The above number represents seconds and we have to convert it in the form of degree, minutes and seconds. To do so, first, we have to determine how many seconds are there in an hour.

1 hour = 3600 seconds

1 minute = 60 seconds

To convert seconds into degree, minutes and seconds, we will divide the seconds by 60 twice.

\begin{array}{ccccccccccc} & & 136&\\ \cline{3-10} \multicolumn{2}{r}{60 \surd} & 8179&& \\ & &8160& & & & \\ \cline{3-7} \cline{6-10} & &19&&\\ \end{array}

The quotient 136 will be further divided by 60 to get hours or degree like this:

\begin{array}{ccccccccccc} & & &\\ \cline{3-10} \multicolumn{2}{r}{60 \surd} & 136&& \\ & &120& & & & \\ \cline{3-7} \cline{6-10} & &16&&\\ \end{array}

The remainder will be minutes and the quotient is hour.

8179'' = 2^o 16' 19''

27,520''

To convert seconds into degree, minutes and seconds, we will divide the seconds by 60 twice.

\begin{array}{ccccccccccc} & & 458&\\ \cline{3-10} \multicolumn{2}{r}{60 \surd} & 27520&& \\ & &27480& & & & \\ \cline{3-7} \cline{6-10} & &40&&\\ \end{array}

 

The remainder 40 is seconds and the quotient 458 will be further divided by 60 to get hours or degree like this:

\begin{array}{ccccccccccc} & & 7&\\ \cline{3-10} \multicolumn{2}{r}{60 \surd} & 458&& \\ & &420& & & & \\ \cline{3-7} \cline{6-10} & &38&&\\ \end{array}

The remainder will be minutes and the quotient represents hour.

27520'' = 7^o 38' 40''

 

Solution of exercise 2

Convert the following units into decimal form:

12° 30' 42''

2h              =  2 x 3600s      =   7200 s

48 min     =   48 x 60          = 2880 s

30 s                                        =  30 s

2 h  48 min 30 s                  = 10110s

 

Solution of exercise 3

Express the following angles in degrees:

3 rad

\frac {\pi}{3} = \frac {180 ^o}{\alpha} = \frac {180^o \cdot 3}{\pi} = 171.887^o = 171^o53'14''

0.887^o . 60 = 53.24'

0.24' \cdot 60 = 14''

 

\frac {2 \pi}{5} rad

= \frac {2.180^o}{5}

= 72^o

 

\frac {3 \pi}{10} rad

= \frac {3.180 ^o}{10} = 54^o

 

Solution of exercise 4

Express the following angles in radians:

316°

\frac {\pi}{\alpha}

\alpha = \frac {316 \pi}{180} = \frac {79\pi}{45} rad

 

10°

\frac {\pi}{\alpha} = \frac {180^o}{10^o}

\alpha = \frac {10 \pi}{180} = \frac {\pi}{18} rad

 

127º

\frac {\pi}{\alpha} = \frac {180^o}{127^o}

\alpha = \frac {127 \pi}{180}

= 2.216 rad

 

Solution of exercise 5

Calculate:

68º 35' 42'' + 56º 46' 39''.

\begin{document} \begin{tabular}{cccc}    & 68^o  &  35'  &  42''\\ + & 56^o  & 46'   & 39'' \\ \hline     &124^o  & 81' & 81'' \\ \end{tabular}

Divide the seconds by 60, if they are greater than 60.

\begin{array}{ccccccccccc} & & 1&\\ \cline{3-10} \multicolumn{2}{r}{60 \surd} & 81&& \\ & &60& & & & \\ \cline{3-7} \cline{6-10} & &21&&\\ \end{array}

124^o 82' 21''

Repeat the same process for minutes

\begin{array}{ccccccccccc} & & 1&\\ \cline{3-10} \multicolumn{2}{r}{60 \surd} & 82&& \\ & &60& & & & \\ \cline{3-7} \cline{6-10} & &22&&\\ \end{array}

125^o 22' 21''

 

Solution of exercise 6

Calculate:

(132° 26' 33'') × 5

= 660^o 130' 165''

Divide 165 by 60:

\begin{array}{ccccccccccc} & & 2&\\ \cline{3-10} \multicolumn{2}{r}{60 \surd} & 165&& \\ & &120& & & & \\ \cline{3-7} \cline{6-10} & &45&&\\ \end{array}

Add the quotient in minutes:

660^o 132'45''

Divide 132 by 60:

\begin{array}{ccccccccccc} & & 2&\\ \cline{3-10} \multicolumn{2}{r}{60 \surd} & 132&& \\ & &120& & & & \\ \cline{3-7} \cline{6-10} & &12&&\\ \end{array}

Add 2 in hours:

662^o 12' 45''

 

(128° 42' 36'') × 3

= 384^o 126' 108''

Divide 108 by 60 like this:

\begin{array}{ccccccccccc} & & 1&\\ \cline{3-10} \multicolumn{2}{r}{60 \surd} & 108&& \\ & &60& & & & \\ \cline{3-7} \cline{6-10} & &48&&\\ \end{array}

= 384^o 127' 48''

Divide 127 by 60:

\begin{array}{ccccccccccc} & & 2&\\ \cline{3-10} \multicolumn{2}{r}{60 \surd} & 127&& \\ & &120& & & & \\ \cline{3-7} \cline{6-10} & &7&&\\ \end{array}

= 386^o 7' 48''

 

Solution of exercise 7

Calculate (132° 26' 33'') : 3

Divide hours by 3 like this:

\begin{array}{ccccccccccc} & & 2&\\ \cline{3-10} \multicolumn{2}{r}{60 \surd} & 132&& \\ & &120& & & & \\ \cline{3-7} \cline{6-10} & &12&&\\ \end{array}

The quotient 2 represents hours and the remainder 12 will be multiplied by 60 and the resulting number will be added to minutes:

26' + 720' = 746'

Repeat the same process for minutes:

\begin{array}{ccccccccccc} & & 248&\\ \cline{3-10} \multicolumn{2}{r}{3\surd} & 746&& \\ & &744& & & & \\ \cline{3-7} \cline{6-10} & &2&&\\ \end{array}

The quotient 248 represents the minutes and the remainder 2 will be multiplied by 60 and the resulting number will be added to seconds:

33'' + 120'' = 153''

Hence, the final answer is:

2^o 248' 153''

 

Solution of exercise 8

Determine the complementary and supplementary angle of 38° 36' 43''.

Complementary angle

We know that the sum of two supplementary angles is equal to 90^o.  So, we will subtract the above angle from 90^o. Because the 90^o is not given in the form of minutes and seconds, so we will subtract 1 from 90^o and place full degree 60' in minutes place. However, we do not have seconds either, so we will subtract 1 again from 60 to get 59'' and place it in the seconds place.

\begin{document} \begin{tabular}{cccc}    & 89^o  &  59'  &  60''\\ - & 38^o  & 36'   & 43'' \\ \hline     &51^o  & 23' & 17'' \\ \end{tabular}

Supplementary angle

The sum of two supplementary angles is 180^o. Hence, we will transform 180^o  to 179^o59'60''.

\begin{document} \begin{tabular}{cccc}    & 179^o  &  59'  &  60''\\ - & 38^o  & 36'   & 43'' \\ \hline     &141^o  & 36' & 43'' \\ \end{tabular}

 

Solution of exercise 9

Determine the complementary and supplementary angle of 25° 38' 40''.

Complementary angle

\begin{document} \begin{tabular}{cccc}    & 89^o  &  59'  &  60''\\ - & 25^o  & 38'   & 40'' \\ \hline     &64^o  & 21' & 20'' \\ \end{tabular}

Hence, the complementary angle is 64^o 21' 20''

 

Supplementary angle

\begin{document} \begin{tabular}{cccc}    & 179^o  &  59'  &  60''\\ - & 25^o  & 38'   & 40'' \\ \hline     &154^o  & 21' & 20'' \\ \end{tabular}

Hence, the supplementary angle is 154^o 21' 20''.

Need a Maths teacher?

Did you like the article?

1 Star2 Stars3 Stars4 Stars5 Stars 3.00/5 - 2 vote(s)
Loading...

Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.