Question

Someone please help me

Answer 1

`\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad \qquad % cosine cos(\theta)=\cfrac{adjacent}{hypotenuse}\\\\ -------------------------------\\\\`


`\bf sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad \qquad sin(\theta_1)=\cfrac{11}{61}\cfrac{\leftarrow opposite=b}{\leftarrow hypotenuse=c} \\\\\\ \textit{so, now we know who is the hypotenuse and opposite side}\\ \textit{let's find the adjacent side then} \\\\\\ c^2=a^2+b^2\implies c^2-b^2=a^2\implies \pm√(c^2-b^2)=a \\\\\\ \pm√(61^2-11^2)=a\implies \pm√(3600)=a\implies \pm 60=a`

but... which is it? the +/-? well, we know the angle is in the first quadrant, in the first quadrant cosine as well as sine, or "a" and "b" are both positive, so then a = 60


`\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\implies cos(\theta_1)=\cfrac{60}{61}`