I am trying to solve this: 3sin2Δ=1.632 for the value of Δ when Δ is between 0 degree and 180 degree. i haven't done it in a long time. can i get the steps please?
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As far as i know its done this way..
3sin2Δ = 1.632
sin2Δ = 1.632 / 3
sin2Δ = .544
2Δ = arcsin(.544)
2Δ = 32.956
Δ = 32.956/2 = 16.478
3sin2Δ = 1.632
sin2Δ = 1.632 / 3
sin2Δ = .544
2Δ = arcsin(.544)
2Δ = 32.956
Δ = 32.956/2 = 16.478