Question

How do i do this problem

Answer 1

`\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad cos(\theta)=\cfrac{adjacent}{hypotenuse}\\\\ -------------------------------\\\\ cos(\theta )=\cfrac{2√(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 √(c^2-a^2)=b\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}`


`\bf \pm \sqrt{7^2-(2√(10))^2}=b\implies \pm\sqrt{49-(2^2√(10^2))}=b \\\\\\ \pm√(49-(4\cdot 10))=b\implies \pm√(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}`