Question

Calculate the angle of (a) $1^0$ (degree) (b) $1'$ (minute of arc or arcmin) and (c) $1''$ (second of arc or arc second)0 in radians. Use $360^0 = 2\pi \:rad, 1^0 =60' \:and\: 1' = 60''$

Answer 1

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1 degree = 60 minute

1 degree = 60 minute1 minute of an arc = 1/60 degree

1 degree = 60 minute1 minute of an arc = 1/60 degree180 degree = pi radian

1 degree = 60 minute1 minute of an arc = 1/60 degree180 degree = pi radian1 degree = pi/180 radian

1 degree = 60 minute1 minute of an arc = 1/60 degree180 degree = pi radian1 degree = pi/180 radiantherefore 1/60 of pi/180 = 1minute

1 degree = 60 minute1 minute of an arc = 1/60 degree180 degree = pi radian1 degree = pi/180 radiantherefore 1/60 of pi/180 = 1minute0r 1 minute = pi /( 60 .180) radian

1 degree = 60 minute1 minute of an arc = 1/60 degree180 degree = pi radian1 degree = pi/180 radiantherefore 1/60 of pi/180 = 1minute0r 1 minute = pi /( 60 .180) radiantherefore 1 min . = 2.9 1.10^-4 radian

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Answer 2

$\text{(a) We have } 360^\circ = 2 \pi \text{ rad}\\\text{ } 1^\circ = (\pi / 180) \text{ rad} = 1.745 \times 10^{-2} \text{ rad}\\\text{(b) } 1^\circ = 60' = 1.745 \times 10^{-2} \text{ rad}\\1' = 2.908 \times 10^{-4} \text{ rad} \:;\: 2.91 \times 10^{-4}\text{ rad}\\\text{(c) } 1' = 60'' = 2.908 \times 10^{-4} \text{ rad}\\1'' = 4.847 \times 10^{-4} \text{ rad} \: ; \: 4.85 \times 10^{-6} \text{ rad}$