Observa as figuras e determina, em cada caso, os valores de x e de y.<\/p>\n
A amplitude de um arco de circunfer\u00eancia \u00e9 igual \u00e0 amplitude do \u00e2ngulo ao centro correspondente. \u00ad<\/p>\n Logo, $y = 360^\\circ\u00a0 – A\\widehat OB = 360^\\circ\u00a0 – 90^\\circ\u00a0 = 270^\\circ $.<\/p>\n \u00ad<\/p>\n Portanto, $x = 40^\\circ $.<\/p>\n \u00ad<\/p>\n $$\\begin{array}{*{20}{l}} Portanto, $x = A\\widehat BC = 40^\\circ $.<\/p>\n \u00ad<\/p>\n Logo, $x = B\\widehat AC = \\frac{{\\mathop {BC}\\limits^\\frown\u00a0 }}{2} = \\frac{{180^\\circ }}{2} = 90^\\circ $.<\/p>\n \u00ad<\/p>\n \u00ad<\/p>\n \u00ad f)<\/strong> Logo, $x = y = D\\widehat AC = D\\widehat BC = \\frac{{\\mathop {CD}\\limits^\\frown\u00a0 }}{2} = \\frac{{100^\\circ }}{2} = 50^\\circ $.<\/p>\n \u00ad<\/p>\n![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/pag23-1.jpg\" "\\"Figura\\"")
<\/a><\/p>\n<\/a>a)<\/strong>
\nComo os \u00e2ngulos AOB e COD s\u00e3o verticalmente opostos, ent\u00e3o s\u00e3o geometricamente iguais.
\nLogo, $y=C\\widehat{O}D=A\\widehat{O}B=30{}^\\text{o}$.<\/p>\n
\nLogo, $x = \\mathop {CD}\\limits^\\frown\u00a0\u00a0 = C\\widehat OD = 30^\\circ $.<\/p>\n<\/a>b)<\/strong>
\nComo o \u00e2ngulo AOB \u00e9 reto, ent\u00e3o $A\\widehat OB = 90^\\circ $.<\/p>\n<\/a>c)<\/strong>
\nComo a amplitude de um arco de circunfer\u00eancia \u00e9 igual \u00e0 amplitude do \u00e2ngulo ao centro correspondente, temos:$$\\begin{array}{*{20}{l}}
\n{x + 30 = 2x – 10}& \\Leftrightarrow &{x – 2x =\u00a0 – 10 – 30} \\\\
\n{}& \\Leftrightarrow &{ – x =\u00a0 – 40} \\\\
\n{}& \\Leftrightarrow &{x = 40}
\n\\end{array}$$<\/p>\n<\/a>d)<\/strong>
\nComo a amplitude de um \u00e2ngulo inscrito \u00e9 igual \u00e0 amplitude do arco compreendido entre os seus lados, temos:<\/p>\n
\n{A\\widehat BC}& = &{\\frac{{\\mathop {AC}\\limits^\\frown\u00a0 }}{2}} \\\\
\n{}& = &{\\frac{{360^\\circ\u00a0 – (\\mathop {AB}\\limits^\\frown\u00a0\u00a0 + \\mathop {BC}\\limits^\\frown\u00a0 )}}{2}} \\\\
\n{}& = &{\\frac{{360^\\circ\u00a0 – (70^\\circ\u00a0 + 130^\\circ )}}{2}} \\\\
\n{}& = &{80^\\circ }
\n\\end{array}$$<\/p>\n<\/a>e)<\/strong>
\nComo [BC] \u00e9 um di\u00e2metro da circunfer\u00eancia, ent\u00e3o o \u00e2ngulo BAC \u00e9 um \u00e2ngulo inscrito num arco de semicircunfer\u00eancia, pelo que \u00e9 reto.<\/p>\n
\n<\/a><\/p>\n
\nOs \u00e2ngulos DAC e DBC s\u00e3o geometricamente iguais, pois s\u00e3o \u00e2ngulos inscritos num mesmo arco de circunfer\u00eancia (arco DAC).<\/p>\n