how do i do this problem

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02-08-2017
Mathematics

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how do i do this problem

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jdoe0001 jdoe0001 · Answer 1 · 2017-08-02T00:51:30+02:00

[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad cos(\theta)=\cfrac{adjacent}{hypotenuse}\\\\ -------------------------------\\\\ cos(\theta )=\cfrac{2\sqrt{10}}{7}\cfrac{\leftarrow adjacent}{\leftarrow hypotenuse}\impliedby \textit{now, let's find the opposite side} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm \sqrt{c^2-a^2}=b\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}[/tex]

[tex]\bf \pm \sqrt{7^2-(2\sqrt{10})^2}=b\implies \pm\sqrt{49-(2^2\sqrt{10^2})}=b \\\\\\ \pm\sqrt{49-(4\cdot 10)}=b\implies \pm\sqrt{9}=b\implies \pm 3=b \\\\\\ \textit{no quadrant is given for the angle, so, we can just assume is the +3} \\\\\\ sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad \qquad sin(\theta )=\cfrac{3}{7}[/tex]