Question

Can you help How can I find sin a I tried but I need help

Answer 1

0 < a < 180/2, is just another way of saying 0° < a < 90°, which is another way of saying angle "a" is in the I Quadrant, where cosine and sine or x,y are both positive.

$\bf cos(a)=\cfrac{\stackrel{adjacent}{5}}{\stackrel{hypotensue}{13}}\impliedby \textit{now let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-5^2}=b\implies \pm\sqrt{144}=b\implies \pm 12=b\implies \stackrel{I~quadrant}{+12=b} \\\\\\ sin(a)=\cfrac{\stackrel{opposite}{12}}{\stackrel{hypotenuse}{13}}$