Anonymous Asked: October 2, 20212021-10-02T06:45:46+05:30 2021-10-02T06:45:46+05:30In: Mathematics Which of the following is correct? a) Sin 1^{0} < Sin 1 b) Sin 1^{0} = Sin 1 c) Sin 1^{0} < Sin 1 d) Sin 2^{0} < Sin 2 1 Answer Voted Oldest Recent Anonymous 2021-10-02T06:51:17+05:30Added an answer on October 2, 2021 at 6:51 am The correct answer is c) Sin 1^{0}Â < Sin 1 where, 180^{0}=Ï€rads. Therefore 1rad=180^{0}â€‹/ Ï€ = 180^{0 / 3.142}â€‹ =57.32^{0} Hence, sin1 =sin 57.32^{0} Thus, sin1^{0}< sin 57.32^{âˆ˜}. Hence, sin1^{0}<sin1. 1 Reply Share Share Share on Facebook Share on Twitter Share on LinkedIn Share on WhatsApp Leave an answerLeave an answerCancel reply Featured image Select file Browse Add a Video to describe the problem better. Video type Youtube Vimeo Dailymotion Facebook Choose from here the video type. Video ID Put Video ID here: https://www.youtube.com/watch?v=sdUUx5FdySs Ex: "sdUUx5FdySs". Answer Anonymously Save my name, email, and website in this browser for the next time I comment. Related Questions How many triangles are there in a pentagram? How many rational numbers are there between two rational numbers? How many significant figures are present in 0.0025? How many 2 digit numbers are divisible by 3? What is the SI unit of inertia? Which value is equal to 1/4 of 1360? What is 78% as a simplified fraction? Which expression is equivalent to (2x^4y)^3

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

^{0}=Ï€rads.Therefore

1rad=180

^{0}â€‹/ Ï€ = 180^{0 / 3.142}â€‹=57.32

^{0}Hence, sin1 =sin 57.32

^{0}Thus, sin1

^{0}< sin 57.32^{âˆ˜}.Hence,

sin1^{0}<sin1.