Question

Express in grades, minutes and seconds, the angles:
(i) ${{138}^{\circ }}30'$
(ii) ${{35}^{\circ }}47'15''$

Answer 1

First of all we will have to convert the given equation completely into degrees by using the conversion $1'={{\left( \dfrac{1}{60} \right)}^{\circ }}$ and $1''={{\left( \dfrac{1}{3600} \right)}^{\circ }}$.
After that we will use the following conversion to convert degrees into grades, minutes and seconds.
${{1}^{\circ }}=\dfrac{10}{9}grade$
${{1}^{\circ }}=60\operatorname{minutes}$
${{1}^{\circ }}=3600\operatorname{seconds}$

Complete step-by-step answer:
The given angles are expressed in grades, minutes and seconds as follows:
(i) ${{138}^{\circ }}30'$
First of all we will convert 30’ into degrees.
We know that $1'={{\dfrac{1}{60}}^{\circ }}$.
$\Rightarrow 30'={{\left( 30\times \dfrac{1}{60} \right)}^{\circ }}={{\dfrac{1}{2}}^{\circ }}={{0.5}^{\circ }}$
$\Rightarrow {{138}^{\circ }}30'={{138}^{\circ }}+30'$
On substituting the value of 30’ we get as follows:
${{138}^{\circ }}30'={{138}^{\circ }}+{{0.5}^{\circ }}={{138.5}^{\circ }}$
We know that ${{1}^{\circ }}=\dfrac{10}{9}grade$.
$\Rightarrow {{138.5}^{\circ }}=\left( 138.5\times \dfrac{10}{9} \right)grade=\dfrac{1385}{9}grades=153.88grade$ (approx.)
Again we know that ${{1}^{\circ }}=60\operatorname{minutes}$.
$\Rightarrow {{138.5}^{\circ }}=\left( 138.5\times 60 \right)\operatorname{minutes}=8310\operatorname{minutes}$
Also, we know that ${{1}^{\circ }}=3600\operatorname{seconds}$.
$\Rightarrow {{138.5}^{\circ }}=\left( 138.5\times 3600 \right)\operatorname{seconds}=498600seconds$
Hence, ${{138}^{\circ }}30'$ is equal to 153.88 grade (approximately) 8310 minutes and 498600 seconds.
(ii) ${{35}^{\circ }}47'15''$
First of all we will have to convert 47’ as well as 15’’ into degrees.
Since we know that $1'={{\dfrac{1}{60}}^{\circ }}$.
$\Rightarrow 47'={{\left( 47\times \dfrac{1}{60} \right)}^{\circ }}={{\dfrac{47}{60}}^{\circ }}$
We also know that $1''={{\dfrac{1}{3600}}^{\circ }}$.
$\Rightarrow 15''={{\left( 15\times \dfrac{1}{3600} \right)}^{\circ }}={{\dfrac{1}{240}}^{\circ }}$
$\Rightarrow {{35}^{\circ }}47'15''={{35}^{\circ }}+47'+15''$
On substituting the values of 47’ and 15’’ we get as follows:
${{35}^{\circ }}47'15''=\dfrac{{{35}^{\circ }}\times 240+47\times 4+1}{240}=\dfrac{8400+188+1}{240}={{\left( \dfrac{2863}{80} \right)}^{\circ }}$
Now, we know that ${{1}^{\circ }}=\dfrac{10}{9}grade$.
$\Rightarrow {{\dfrac{2863}{80}}^{\circ }}=\left( \dfrac{2863}{80}\times \dfrac{10}{9} \right)grade=\dfrac{2863}{72}grade=39.76grade$ (approx.)
Also, we know that ${{1}^{\circ }}=60\operatorname{minutes}$.
$\Rightarrow {{\left( \dfrac{2863}{80} \right)}^{\circ }}=\left( \dfrac{2863}{80}\times 60 \right)\operatorname{minutes}=\dfrac{8589}{4}\operatorname{minutes}$
Once again, we know that ${{1}^{\circ }}=3600\operatorname{seconds}$.
$\Rightarrow {{\left( \dfrac{2863}{80} \right)}^{\circ }}=\left( \dfrac{2863}{80}\times 3600 \right)\operatorname{seconds}=128835seconds$
Hence the angle ${{35}^{\circ }}47'15''$ is equal to 39.76 grade (approximately) $\dfrac{8589}{4}$ minutes and 128835 seconds.
Therefore, the given angles are expressed in grades, minutes and seconds.

Note: While solving this question, there is a chance that we might write ${{138}^{\circ }}30'={{138}^{\circ }}\times {{0.5}^{\circ }}={{69}^{\circ }}$ instead of ${{138}^{\circ }}30'={{138}^{\circ }}+{{0.5}^{\circ }}={{138.5}^{\circ }}$, so we must remember that if $b'={{c}^{\circ }}$, then ${{a}^{\circ }}b'={{a}^{\circ }}+{{c}^{\circ }}$ and not ${{a}^{\circ }}b'={{a}^{\circ }}\times {{c}^{\circ }}$. Sometimes we use ${{1}^{\circ }}=\dfrac{9}{10}grade$ by mistake which is wrong. So, it must be avoided. Also, remember that the given angle must be completely into degree or radian then we will use the conversion according to it. If the given angle is in the form of ${{x}^{\circ }}y'z''$ then first of all we will convert minutes and seconds into degrees.