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If tan θ = ¾, then Sin θ is:
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tan = opposite / adjacent sin = opposite / hypotenuse u have 2 draw a triangle... right angled... use Pythagoras theorem to solve for the unknown side ***************************************************************************
If tangent of theta = 3/4 then theta = arctangent of 3/4 or 36.87 Degrees. So the sine of theta = sine 36.87 Degrees = .6
or u can do like this.. \[\sin^2 \theta =1-\cos^2 \theta=1-\frac{ 1 }{ \sec^2 \theta }=1-\frac{ 1 }{ 1+ \tan^2 \theta }\] then square root it
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