I have a test tomorrow :(
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if Θ = π/10, then
tan(Θ) = tan(π/10),
and since tan(Θ) = y/x,
y=[tan(π/10)]∙x
use calculator for approximation for tan(π/10)
hope that helps.
tan(Θ) = tan(π/10),
and since tan(Θ) = y/x,
y=[tan(π/10)]∙x
use calculator for approximation for tan(π/10)
hope that helps.
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tanΘ = y/x provided x ≠ 0.
So
y/x = tan(π/10) ==> y = tan(π/10)x.
This is a line through the origin with slope m = tan(π/10).
So
y/x = tan(π/10) ==> y = tan(π/10)x.
This is a line through the origin with slope m = tan(π/10).