a) Sin 1^{0} < Sin 1

b) Sin 1^{0} = Sin 1

c) Sin 1^{0} < Sin 1

d) Sin 2^{0} < Sin 2

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The correct answer is c) Sin 1^{0}< Sin 1where, 180

^{0}=πrads.Therefore

1rad=180

^{0}/ π = 180^{0 / 3.142}=57.32

^{0}Hence, sin1 =sin 57.32

^{0}Thus, sin1

^{0}< sin 57.32^{∘}.Hence,

sin1^{0}<sin1.