\nDesenhado esse \u00e2ngulo com o aux\u00edlio do c\u00edrculo trigonom\u00e9trico, o seu lado extremidade intersecta a circunfer\u00eancia no ponto $P\\,(-\\frac{3}{4},y)$, com $y>0$. (Porqu\u00ea?)<\/p>\n
Como $\\overline{OP}=1$,\u00a0temos: \\[\\left\\{ \\begin{array}{*{35}{l}}
\n\\sqrt{{{(-\\frac{3}{4}-0)}^{2}}+{{(y-0)}^{2}}}=1\u00a0 \\\\
\ny>0\u00a0 \\\\
\n\\end{array} \\right.\\Leftrightarrow \\left\\{ \\begin{array}{*{35}{l}}
\n\\frac{9}{16}+{{y}^{2}}=1\u00a0 \\\\
\ny>0\u00a0 \\\\
\n\\end{array} \\right.\\Leftrightarrow \\left\\{ \\begin{array}{*{35}{l}}
\n{{y}^{2}}=\\frac{7}{16}\u00a0 \\\\
\ny>0\u00a0 \\\\
\n\\end{array} \\right.\\Leftrightarrow \\left\\{ \\begin{array}{*{35}{l}}
\ny=\\mp \\frac{\\sqrt{7}}{4}\u00a0 \\\\
\ny>0\u00a0 \\\\
\n\\end{array} \\right.\\Leftrightarrow y=\\frac{\\sqrt{7}}{4}\\]
\nAssim, \\(P\\,(-\\frac{3}{4},\\frac{\\sqrt{7}}{4})\\).
\nLogo, \\[sen\\,\\alpha =\\frac{\\sqrt{7}}{4}\\ \\ \\ \\text{e}\\ \\ \\ tg\\,\\alpha =\\frac{\\frac{\\sqrt{7}}{4}}{-\\frac{3}{4}}=-\\frac{\\sqrt{7}}{3}\\]<\/p>\n