
Express in terms of a right angle, the angle \[{{63}^{\circ }}17'25''\].
Answer
593.4k+ views
Hint: We know that the right angle is equal to \[{{90}^{\circ }}\]. First of all we will convert the given angle completely into degrees which means we have to convert minutes and seconds into degrees for which we have the following conversions:
\[\begin{align}
& 1'={{\left( \dfrac{1}{60} \right)}^{\circ }} \\
& 1''={{\left( \dfrac{1}{3600} \right)}^{\circ }} \\
\end{align}\]
After converting the given angle completely into degrees, we will suppose the given angle is equal to ‘k’ times the right angle and thus equating them we will find the value of ‘k’.
Complete step-by-step answer:
We have been given the angle \[{{63}^{\circ }}17'25''\] which we have to express in terms of a right angle.
We know that \[1'={{\dfrac{1}{60}}^{\circ }}\]
\[\Rightarrow 17'={{\left( 17\times \dfrac{1}{60} \right)}^{\circ }}={{\dfrac{17}{60}}^{\circ }}\]
Also, we know that \[1''={{\left( \dfrac{1}{3600} \right)}^{\circ }}\]
\[25''={{\left( 25\times \dfrac{1}{3600} \right)}^{\circ }}={{\left( \dfrac{25}{3600} \right)}^{\circ }}={{\left( \dfrac{1}{144} \right)}^{\circ }}\]
\[\Rightarrow {{63}^{\circ }}17'25''={{63}^{\circ }}+17'+25''\]
Now substituting the values of 17’ and 25’’.
\[{{63}^{\circ }}17'25''=63+\dfrac{17}{60}+\dfrac{1}{144}\]
Now taking LCM of 60 and 144, we get as follows:
\[{{63}^{\circ }}17'25''=\dfrac{720\times 63+17\times 12+5}{720}=\dfrac{45360+204+5}{720}=\dfrac{{{45569}^{\circ }}}{720}\]
Let us suppose \[\dfrac{{{45569}^{\circ }}}{720}\] is equal to k times of a right angle.
\[\Rightarrow \dfrac{{{45569}^{\circ }}}{720}=k\times {{90}^{\circ }}\]
On dividing by \[{{90}^{\circ }}\] to both sides of equation, we get as follows:
\[\begin{align}
& \Rightarrow \dfrac{{{45569}^{\circ }}}{720\times 90}=\dfrac{k\times 90'}{90'} \\
& \Rightarrow k=\dfrac{45569}{64800} \\
\end{align}\]
Therefore, \[{{63}^{\circ }}17'25''\] can be expressed in terms of right angle as \[\dfrac{45569}{64800}\] times a right angle.
Note: In this type of questions, first of all check that the given angle is completely either in degree or radians then suppose the angle to be equal to k times right angle and the unit of measurement must be the same. If the given angle is in degrees then you will take right angle equal to \[{{90}^{\circ }}\] and if the given angle is in radians then you will take right angle equal to \[\dfrac{\pi }{2}\] radians. Also, sometimes in order to calculate the value of ‘k’ we just divide by \[{{90}^{\circ }}\] by the given angle by mistake. So, be careful while calculating the value of k.
\[\begin{align}
& 1'={{\left( \dfrac{1}{60} \right)}^{\circ }} \\
& 1''={{\left( \dfrac{1}{3600} \right)}^{\circ }} \\
\end{align}\]
After converting the given angle completely into degrees, we will suppose the given angle is equal to ‘k’ times the right angle and thus equating them we will find the value of ‘k’.
Complete step-by-step answer:
We have been given the angle \[{{63}^{\circ }}17'25''\] which we have to express in terms of a right angle.
We know that \[1'={{\dfrac{1}{60}}^{\circ }}\]
\[\Rightarrow 17'={{\left( 17\times \dfrac{1}{60} \right)}^{\circ }}={{\dfrac{17}{60}}^{\circ }}\]
Also, we know that \[1''={{\left( \dfrac{1}{3600} \right)}^{\circ }}\]
\[25''={{\left( 25\times \dfrac{1}{3600} \right)}^{\circ }}={{\left( \dfrac{25}{3600} \right)}^{\circ }}={{\left( \dfrac{1}{144} \right)}^{\circ }}\]
\[\Rightarrow {{63}^{\circ }}17'25''={{63}^{\circ }}+17'+25''\]
Now substituting the values of 17’ and 25’’.
\[{{63}^{\circ }}17'25''=63+\dfrac{17}{60}+\dfrac{1}{144}\]
Now taking LCM of 60 and 144, we get as follows:
\[{{63}^{\circ }}17'25''=\dfrac{720\times 63+17\times 12+5}{720}=\dfrac{45360+204+5}{720}=\dfrac{{{45569}^{\circ }}}{720}\]
Let us suppose \[\dfrac{{{45569}^{\circ }}}{720}\] is equal to k times of a right angle.
\[\Rightarrow \dfrac{{{45569}^{\circ }}}{720}=k\times {{90}^{\circ }}\]
On dividing by \[{{90}^{\circ }}\] to both sides of equation, we get as follows:
\[\begin{align}
& \Rightarrow \dfrac{{{45569}^{\circ }}}{720\times 90}=\dfrac{k\times 90'}{90'} \\
& \Rightarrow k=\dfrac{45569}{64800} \\
\end{align}\]
Therefore, \[{{63}^{\circ }}17'25''\] can be expressed in terms of right angle as \[\dfrac{45569}{64800}\] times a right angle.
Note: In this type of questions, first of all check that the given angle is completely either in degree or radians then suppose the angle to be equal to k times right angle and the unit of measurement must be the same. If the given angle is in degrees then you will take right angle equal to \[{{90}^{\circ }}\] and if the given angle is in radians then you will take right angle equal to \[\dfrac{\pi }{2}\] radians. Also, sometimes in order to calculate the value of ‘k’ we just divide by \[{{90}^{\circ }}\] by the given angle by mistake. So, be careful while calculating the value of k.
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