Question

Express in terms of right angles, and also in degrees, minutes and seconds, the angles:
(i) ${{39}^{g}}45'36''$
(ii) ${{255}^{g}}8'9''$
(iii) ${{759}^{g}}0'5''$

Answer 1

Hint: For the above question, we will have to know the conversion of grade, minutes and seconds into degrees as follows:
${{1}^{g}}\left( grade \right)={{\dfrac{9}{10}}^{\circ }}$
$1'\left( \operatorname{minute} \right)={{\dfrac{1}{60}}^{\circ }}$ or ${{1}^{\circ }}=60\operatorname{minutes}$
$1''\left( \operatorname{seconds} \right)={{\dfrac{1}{3600}}^{\circ }}$ or ${{1}^{\circ }}=3600\operatorname{seconds}$

Complete step-by-step answer:
We also know that the right angle is equal to ${{90}^{\circ }}$ . 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’.
The given angles are expressed in terms of right angles and also in degrees, minutes and seconds as follows:
(i) ${{39}^{g}}45'36''$
Since we know that ${{1}^{g}}\left( grade \right)={{\dfrac{9}{10}}^{\circ }}$.
$\Rightarrow 39grade={{\left( 39\times \dfrac{9}{10} \right)}^{\circ }}={{\left( \dfrac{351}{10} \right)}^{\circ }}$.
We also know that $1'={{\dfrac{1}{60}}^{\circ }}$.
$\Rightarrow 45'={{\left( 45\times \dfrac{1}{60} \right)}^{\circ }}={{\left( \dfrac{3}{4} \right)}^{\circ }}$
Again we know that $1''={{\dfrac{1}{3600}}^{\circ }}$.
$\Rightarrow 36''={{\left( 36\times \dfrac{1}{3600} \right)}^{\circ }}={{\left( \dfrac{1}{100} \right)}^{\circ }}$
$\Rightarrow {{39}^{g}}45'36''={{\left( \dfrac{351}{10} \right)}^{\circ }}+{{\left( \dfrac{3}{4} \right)}^{\circ }}+{{\left( \dfrac{1}{100} \right)}^{\circ }}$
On taking LCM of 10, 4 and 100, we get as follows:
$\Rightarrow {{39}^{g}}45'36''={{\left( \dfrac{351\times 100+3\times 25+1}{100} \right)}^{\circ }}={{\left( \dfrac{3510+75+1}{100} \right)}^{\circ }}={{\left( \dfrac{3586}{100} \right)}^{\circ }}={{\left( \dfrac{1793}{50} \right)}^{\circ }}$
Let us suppose ${{\left( \dfrac{1793}{50} \right)}^{\circ }}$ to be equal to k times a right angle
$\Rightarrow {{\left( \dfrac{1793}{50} \right)}^{\circ }}=k\times {{90}^{\circ }}$
On dividing by ${{90}^{\circ }}$ on both the sides, we get as follows:

& \Rightarrow \dfrac{1793}{50\times 90}=\dfrac{k\times 90}{90} \\\ & \Rightarrow k=\dfrac{1793}{4500} \\\ \end{aligned}$$ Now we know that $${{1}^{\circ }}=\dfrac{1}{60}\operatorname{minutes}$$. $$\Rightarrow {{\dfrac{1793}{50}}^{\circ }}=\dfrac{1793}{50}\times \dfrac{1}{60}=\dfrac{1793}{3000}\operatorname{minutes}$$ We also know that $$1'=\dfrac{1}{3600}seconds$$. $$\Rightarrow {{\left( \dfrac{1793}{50} \right)}^{\circ }}=\dfrac{1793}{50}\times \dfrac{1}{3600}=\dfrac{1793}{180000}\operatorname{seconds}$$ Hence $${{39}^{g}}45'36''={{\left( \dfrac{1793}{50} \right)}^{\circ }}=\dfrac{1793}{4500}$$ of right angles $$=\dfrac{1793}{3000}$$ minutes $$=\dfrac{1793}{180000}$$ seconds. (ii) $${{255}^{g}}8'9''$$ Since we know that $${{1}^{g}}\left( grade \right)={{\dfrac{9}{10}}^{\circ }}$$. $$\Rightarrow 255grade={{\left( 255\times \dfrac{9}{10} \right)}^{\circ }}={{\left( \dfrac{459}{2} \right)}^{\circ }}$$ We also know that $$1'={{\dfrac{1}{60}}^{\circ }}$$. $$\Rightarrow 8'={{\left( 8\times \dfrac{1}{60} \right)}^{\circ }}={{\dfrac{2}{15}}^{\circ }}$$ Again, we know that $$1''={{\dfrac{1}{3600}}^{\circ }}$$. $$\Rightarrow 9''={{\left( 9\times \dfrac{1}{3600} \right)}^{\circ }}={{\dfrac{1}{400}}^{\circ }}$$ $$\Rightarrow {{255}^{g}}8'9''={{255}^{g}}+8'+9''={{\dfrac{459}{2}}^{\circ }}+{{\dfrac{2}{15}}^{\circ }}+{{\dfrac{1}{400}}^{\circ }}$$ On taking LCM of 2, 15 and 400 we get as follows: $$\Rightarrow \dfrac{459\times 600+2\times 80+3}{1200}=\dfrac{275400+160+3}{1200}={{\left( \dfrac{275563}{1200} \right)}^{\circ }}$$ Let us suppose $${{\left( \dfrac{275563}{1200} \right)}^{\circ }}$$ to be equal to k times a right angle and we know that a right angle is equal to $${{90}^{\circ }}$$. $$\Rightarrow {{\left( \dfrac{275563}{1200} \right)}^{\circ }}=k\times {{90}^{\circ }}$$ On dividing the equation by $$90$$ on both the sides we get as follows: $$\begin{aligned} & \Rightarrow \dfrac{275563}{1200\times 90}=\dfrac{k\times 90}{90} \\\ & \Rightarrow k=\dfrac{275563}{108000} \\\ \end{aligned}$$ Now we know that $${{1}^{\circ }}=60\operatorname{minutes}$$. $$\Rightarrow {{\dfrac{275563}{1200}}^{\circ }}=\dfrac{275563}{1200}\times 60=\dfrac{275563}{20}\operatorname{minutes}$$ We also know that $${{1}^{\circ }}=3600\operatorname{seconds}$$. $$\Rightarrow {{\dfrac{275563}{1200}}^{\circ }}=\dfrac{275563}{1200}\times 3600=826689\operatorname{seconds}$$ Hence $${{255}^{g}}8'9''={{\left( \dfrac{275563}{1200} \right)}^{\circ }}=\dfrac{275563}{108000}$$ times a right angle $$=\dfrac{275563}{20}\operatorname{minutes}=826689seconds$$ (iii) $${{759}^{g}}0'5''$$ Since $$1grade={{\dfrac{9}{10}}^{\circ }}$$ $$\Rightarrow 759grade={{\left( 759\times \dfrac{9}{10} \right)}^{\circ }}={{\left( \dfrac{6831}{10} \right)}^{\circ }}$$ Also, $$0'={{0}^{\circ }}$$ And we know that $$1''={{\dfrac{1}{3600}}^{\circ }}$$ $$\Rightarrow 5''={{\left( 5\times \dfrac{1}{3600} \right)}^{\circ }}={{\left( \dfrac{1}{720} \right)}^{\circ }}$$ $$\Rightarrow {{759}^{g}}0'5''={{759}^{g}}+0'+5''={{\dfrac{6831}{10}}^{\circ }}+{{0}^{\circ }}+{{\dfrac{1}{720}}^{\circ }}$$ On taking LCM of 10 and 720 we get as follows: $$\Rightarrow \dfrac{6831\times 72+0+1}{720}=\dfrac{491833}{720}$$ Let us suppose $${{\left( \dfrac{491833}{720} \right)}^{\circ }}$$ is equal to k times a right angle and we know that right angle and we know that right angle equals to 90 degrees. $$\Rightarrow {{\dfrac{491833}{720}}^{\circ }}=k\times {{90}^{\circ }}$$ On dividing the equation by $${{90}^{\circ }}$$ on both the sides, we get as follows: $$\begin{aligned} & \Rightarrow \dfrac{491833}{720\times 90}=\dfrac{k\times 90}{90} \\\ & \Rightarrow k=\dfrac{491833}{64800} \\\ \end{aligned}$$ Also, we know that $${{1}^{\circ }}=60\operatorname{minutes}$$. $$\Rightarrow \dfrac{491833}{720}=\dfrac{491833}{720}\times 60\operatorname{minutes}=\dfrac{491833}{12}\operatorname{minutes}$$ Once again, we know that $${{1}^{\circ }}=3600\operatorname{seconds}$$. $$\Rightarrow {{\left( \dfrac{491833}{720} \right)}^{\circ }}=\dfrac{491833}{720}\times 3600=2459165\operatorname{seconds}$$ Hence $${{759}^{g}}0'5''=\left( \dfrac{491833}{720} \right)$$ degrees $$=\dfrac{491833}{64800}$$ of a right angle $$=\dfrac{491833}{12}\operatorname{minutes}$$ $$=2459165seconds$$. Note: Sometimes we use $${{1}^{\circ }}=\dfrac{9}{10}grade$$ by mistake which is wrong. So, be careful while using it. Also, remember that the given angle must be completely into degree or radian then we will use the conversion according to it. If the given angle is in the form of $${{x}^{\circ }}y'z''$$ then first of all we will convert minutes and seconds into degrees. Also remember that a right angle is equal to 90 degrees and $$\dfrac{\pi }{2}$$ radians in degree and radian units of angles respectively.