Relation between Degrees Radians and Grades

In this section we will discuss about the relation between degrees radians and grades.
Let the degree be denoted by 'D', 'R' be the number of radians and 'G' be the number of grades in an angle $\Theta$.
We know that, $90^{0}$ = 1 right angle

⇒ $1^{0} = \frac{1}{90}$ right angle

⇒ $D^{0} = \frac{D}{90}$ right angles

⇒ $\Theta = \frac{D}{90}$ right angle -----------(i)

Again we know that $\pi$ radians = 2 right angles = $180^{0}$

⇒ $1^{c} = \frac{2}{\pi}$ right angle

⇒ R radians = $\frac{2R}{\pi}$ right angles ----------(ii)

And now, 100 grades = 1 right angle

⇒ 1 grade = = $\frac{1}{100}$ right angle

⇒ G grade = $\frac{G}{100}$ right angles

⇒ $\Theta = \frac{G}{100}$ right angles --------(iii)

Thus, from (i), (ii) and (iii), we get,

$\frac{D}{90} = \frac{G}{100} = \frac{2R}{\pi}$

Some useful points on relation between degrees radians and grades

(i) The angle between two consecutive digits in a clock is $30^{0}$. In radians it is $\frac{\pi}{6}$
(ii) The hour hand rotates through an angle of $30^{0}$ in one hour which is $\left ( \frac{1}{2} \right )^{0}$ in one minute.
(iii) The minute hand rotates through an angle of $6^{0}$ in one minute.
Example :
One angle of a triangle is $\left ( \frac{2x}{3} \right )$ grades = $\left ( \frac{2x}{3} \right )^{g}$ and another is $\left ( \frac{3x}{2} \right )^{0}$ while the third is $\left ( \frac{\pi x}{75} \right )$ radians. Express all the angles in degrees.
Solution : We know that ,
$\frac{D}{90} = \frac{G}{100} = \frac{2R}{\pi}$

⇒ $\left ( \frac{2x}{3} \right )^{g} = \left ( \frac{2}{3} x\times\frac{90}{100} \right )^{0} = \left ( \frac{3x}{5} \right )^{0}$ ----(i)

⇒ $\left ( \frac{\pi x}{75} \right )^{c} = \left ( \frac{\pi}{75} x\times\frac{180}{\pi} \right )^{c} = \left ( \frac{12x}{5} \right )^{0}$ ----(ii)

Sum of all angles in a triangle is $180^{0}$
∴ $ \left ( \frac{3}{5}x \right )^{0} + \left ( \frac{3}{2}x \right )^{0} + \left ( \frac{12x}{5} \right )^{0} = 180^{0}$

⇒ x = $40^{0}$
So each angle will be
$ \left ( \frac{3}{5}x \right )^{0} = \left ( \frac{3}{5}\times 40 \right )^{0} = 24^{0}$

$ \left ( \frac{3}{2}x \right )^{0} = \left ( \frac{3}{2}\times 40 \right )^{0} = 60^{0}$

$ \left ( \frac{12}{5}x \right )^{0} = \left ( \frac{12}{5}\times 40 \right )^{0} = 96^{0}$

So the required angles are $24^{0}, 60^{0},96^{0}$


11th grade math

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