Find the zero of the linear function f(x)=x+5
1 answer:
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Cos(<span>θ) < 0, so we know it would be in Quadrant 2 or 3
then csc(</span>θ) = 257, but csc(θ) =
= 257
==> sin(<span>θ) =
it is positive, so now we can determine that is in Quadrant 2
sin(</span>θ) = opp./hyp both of opp and hyp are positive but adj suppose to negative because that way it leads the cos(<span>θ) < 0
</span>cos(<span>θ) = adj/hyp
</span>Pythereom to find the adj:
cosθ =
tanθ =
cos<span>θ = </span>
then csc(</span>θ) = 257, but csc(θ) =
==> sin(<span>θ) =
it is positive, so now we can determine that is in Quadrant 2
sin(</span>θ) = opp./hyp both of opp and hyp are positive but adj suppose to negative because that way it leads the cos(<span>θ) < 0
</span>cos(<span>θ) = adj/hyp
</span>Pythereom to find the adj:
cosθ =
tanθ =
cos<span>θ = </span>