Question

Reduce the degree of the form ${{63}^{\text{o}}}14'51''$ into centesimal measure.

Answer 1

Hint: We will apply the formula used for defining the relationship between degrees and grades. The formula is given by ${{90}^{\text{o}}}={{100}^{g}}$. Here $^{\text{o}}$ is called the degree and g is called grades. We will apply the concept of degree, minutes and seconds which have numerical form as ${{\left( 1 \right)}^{\text{o}}}=60'$ or $\left( 1 \right)'={{\left( \dfrac{1}{60} \right)}^{\text{o}}}$ and ${{\left( 1 \right)}^{\text{o}}}={{\left( \dfrac{1}{3600} \right)}^{\text{o}}}$.

Complete step-by-step answer:
Now, we will consider ${{63}^{\text{o}}}14'51''$. We can write ${{63}^{\text{o}}}14'51''$ as ${{63}^{\text{o}}}+14'+51''$. Here, we will apply the concept of degree, minutes and seconds. As we know that ${{\left( 1 \right)}^{\text{o}}}=60'$ or $\left( 1 \right)'={{\left( \dfrac{1}{60} \right)}^{\text{o}}}$ and ${{\left( 1 \right)}^{\text{o}}}={{\left( \dfrac{1}{3600} \right)}^{\text{o}}}$. Therefore, we have
$\begin{aligned} & {{63}^{\text{o}}}14'51''={{63}^{\text{o}}}+14'+51'' \\\ & \Rightarrow {{63}^{\text{o}}}14'51''={{63}^{\text{o}}}+\left( 14\times 1' \right)+\left( 51\times 1'' \right) \\\ & \Rightarrow {{63}^{\text{o}}}14'51''={{63}^{\text{o}}}+\left( 14\times {{\left( \dfrac{1}{60} \right)}^{\text{o}}} \right)+\left( 51\times {{\left( \dfrac{1}{3600} \right)}^{\text{o}}} \right) \\\ & \Rightarrow {{63}^{\text{o}}}14'51''={{63}^{\text{o}}}+{{\left( \dfrac{14}{60} \right)}^{\text{o}}}+{{\left( \dfrac{51}{3600} \right)}^{\text{o}}} \\\ & \Rightarrow {{63}^{\text{o}}}14'51''={{\left( 63+\dfrac{14}{60}+\dfrac{51}{3600} \right)}^{\text{o}}} \\\ & \Rightarrow {{63}^{\text{o}}}14'51''={{\left( 63+\dfrac{7}{30}+\dfrac{17}{1200} \right)}^{\text{o}}} \\\ \end{aligned}$
As we know that the l.c.m. of 1, 30 and 1200 is 1200. Therefore, we have
$\begin{aligned} & {{63}^{\text{o}}}14'51''={{\left( 63+\dfrac{7}{30}+\dfrac{17}{1200} \right)}^{\text{o}}} \\\ & \Rightarrow {{63}^{\text{o}}}14'51''={{\left( \dfrac{75600+280+17}{1200} \right)}^{\text{o}}} \\\ & \Rightarrow {{63}^{\text{o}}}14'51''={{\left( \dfrac{75897}{1200} \right)}^{\text{o}}} \\\ & \Rightarrow {{63}^{\text{o}}}14'51''={{63.2475}^{\text{o}}} \\\ \end{aligned}$
Now we will convert the degree into grades by the formula which is given by ${{90}^{\text{o}}}={{100}^{g}}$ or, ${{1}^{\text{o}}}={{\left( \dfrac{100}{90} \right)}^{g}}$. Here $^{\text{o}}$ is called the degree and g is called grades. As we know that ${{63.2475}^{\text{o}}}$ can be written as ${{63.2475}^{\text{o}}}=63.2475\times {{1}^{\text{o}}}$. Therefore, after using the formula ${{1}^{\text{o}}}={{\left( \dfrac{100}{90} \right)}^{g}}$ we will get
$\begin{aligned} & {{63.2475}^{\text{o}}}=63.2475\times {{1}^{\text{o}}} \\\ & \Rightarrow {{63.2475}^{\text{o}}}=63.2475\times {{\left( \dfrac{100}{90} \right)}^{g}} \\\ & \Rightarrow {{63.2475}^{\text{o}}}={{\left( 63.2475\times \dfrac{100}{90} \right)}^{g}} \\\ & \Rightarrow {{63.2475}^{\text{o}}}={{\left( \dfrac{63.2475}{90}\times 100 \right)}^{g}} \\\ & \Rightarrow {{63.2475}^{\text{o}}}={{\left( 0.70275\times 100 \right)}^{g}} \\\ & \Rightarrow {{63.2475}^{\text{o}}}={{\left( 70.275 \right)}^{g}} \\\ \end{aligned}$
Now, we will write ${{\left( 70.275 \right)}^{g}}$ in terms of its grades, minutes and seconds. Clearly decimal is after 70 so we have ${{70}^{g}}$ for sure. Now after decimal we have 275 or since it is after decimal so one 0 can be added after it. Therefore, we have 0.275 as 0.2750. And this can be separated as 27 minutes and 50 seconds. Therefore, we can write ${{\left( 70.275 \right)}^{g}}$ as ${{70}^{g}}27'50''$. So, our degree ${{63}^{\text{o}}}14'51''$ is changed into grades as ${{\left( 70.275 \right)}^{g}}$ which is further converted into minutes and seconds as ${{70}^{g}}27'50''$.
Hence, ${{63}^{\text{o}}}14'51''$ can be written as ${{\left( 70.275 \right)}^{g}}$ or ${{70}^{g}}27'50''$.

Note: Alternatively we could have used the value ${{\left( 1 \right)}^{\text{o}}}={{1.111111}^{g}}$ in ${{63.2475}^{\text{o}}}=63.2475\times {{1}^{\text{o}}}$. Therefore, we directly get
$\begin{aligned} & {{63.2475}^{\text{o}}}=63.2475\times {{1}^{\text{o}}} \\\ & \Rightarrow {{63.2475}^{\text{o}}}=63.2475\times {{1.111111}^{g}} \\\ & \Rightarrow {{63.2475}^{\text{o}}}={{70.27499}^{g}} \\\ \end{aligned}$
And approximately we get, ${{63}^{\text{o}}}14'51''$ can be written as ${{70}^{g}}27'50''$.