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[how do you]Find sin theta if tan theta = 3/4

User Alficles
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\bf tan(\theta )=\cfrac{\stackrel{opposite}{3}}{\stackrel{adjacent}{4}}\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies c=√(a^2+b^2) \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=√(3^2+4^2)\implies c=5 \\\\\\ sin(\theta )=\cfrac{\stackrel{opposite}{3}}{\stackrel{hypotenuse}{5}}

bear in mind that, though the square has two valid roots, one negative and one positive, the hypotenuse is just a radius distance, and therefore is never negative.
User Avrohom
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