Asked by Calie
Suppose that sin(t) = x/(x + 3)
Use this information and a reference triangle to eliminate the trigonometric function from the following expression.
inverse tangent (x/ sqrt(6x + 9))
Use this information and a reference triangle to eliminate the trigonometric function from the following expression.
inverse tangent (x/ sqrt(6x + 9))
Answers
Answered by
oobleck
opposite leg is x
hypotenuse is x+3
so adjacent leg is √((x+3)^2 - x^2) = √(6x+9)
so tan(t) = opposite/adjacent = x/√(6x+9)
hypotenuse is x+3
so adjacent leg is √((x+3)^2 - x^2) = √(6x+9)
so tan(t) = opposite/adjacent = x/√(6x+9)
Answered by
Anonymous
tan a = (x/ sqrt(6x + 9))
draw right triangle , legs x and sqrt(6x+9)
hypotenuse = sqrt (x^2 + 6 x + 9) = sqrt (x+3)^2 = x+3
then sin a = x/(x+3)
draw right triangle , legs x and sqrt(6x+9)
hypotenuse = sqrt (x^2 + 6 x + 9) = sqrt (x+3)^2 = x+3
then sin a = x/(x+3)
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