if G,D and R denotes respectively the no. of grades , degrees and radians in an angle , prove that G-D=20R/pi.

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Solution

$G^{g} = D^{0} = R radian \frac{G}{100} right angle = \frac{D}{90} right angle = \frac{R}{\frac{π}{2}} right angle \frac{G}{100} = \frac{2 R}{π} and \frac{D}{90} = \frac{2 R}{π} So G = \frac{200 R}{π} and D = \frac{180 R}{π} SO G - D = \frac{20 R}{π}$

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If D, G and R denote respectively the number of degrees, grades and radians in an angle, then

Q. If D, G and R denote respectively the number of degrees, grades and radians in an angle, then
(a)

\frac{D}{100} = \frac{G}{90} = \frac{2 R}{π}

(b)

\frac{D}{90} = \frac{G}{100} = \frac{R}{π}

(c)

\frac{D}{100} = \frac{G}{100} = \frac{2 R}{π}

(d)

\frac{D}{90} = \frac{G}{100} = \frac{R}{2 π}

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Q. The measure of an angle in degrees, grades and radians be D, G and C respectively, then relation between them

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if G,D and R denotes respectively the no. of grades , degrees and radians in an angle , prove that G-D=20R/pi.

Gg=D0=R radianG100right angle =D90right angle =Rπ2right angleG100=2Rπ and D90=2RπSo G=200Rπ and D=180RπSO G-D=20Rπ

$G^{g} = D^{0} = R radian \frac{G}{100} right angle = \frac{D}{90} right angle = \frac{R}{\frac{π}{2}} right angle \frac{G}{100} = \frac{2 R}{π} and \frac{D}{90} = \frac{2 R}{π} So G = \frac{200 R}{π} and D = \frac{180 R}{π} SO G - D = \frac{20 R}{π}$