\nEnunciado<\/b><\/span><\/p>\n
Observa as figuras e calcula, em cada caso, o valor de x.<\/p>\n\n\n\n
$\"\"$ <\/a><\/td>\n $\"\"$ <\/a><\/td>\n $\"\"$ <\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
Resolu\u00e7\u00e3o >><\/a><\/span><\/div><\/div>\n\n
\nResolu\u00e7\u00e3o<\/b><\/span><\/p>\n
$\"\"$ <\/a>O \u00e2ngulo CAD \u00e9 um \u00e2ngulo externo do tri\u00e2ngulo [ADM], pelo que a sua amplitude \u00e9 igual \u00e0 soma das amplitudes dos \u00e2ngulos internos n\u00e3o adjacentes.<\/p>\n
Isto \u00e9: $$\\begin{array}{*{20}{l}}
\n{C\\widehat AD}& = &{A\\widehat DM + A\\widehat MD} \\\\
\n{}& = &{14^\\circ\u00a0 + 20^\\circ } \\\\
\n{}& = &{34^\\circ }
\n\\end{array}$$<\/p>\n
Como o \u00e2ngulo CAD \u00e9 um \u00e2ngulo inscrito, a amplitude do arco compreendido entre os seus lados \u00e9 dupla da amplitude deste \u00e2ngulo inscrito. Assim, $\\mathop {CD}\\limits^\\frown\u00a0\u00a0 = 2 \\times C\\widehat AD = 2 \\times 34^\\circ\u00a0 = 68^\\circ $.<\/p>\n
Logo, $x = 68^\\circ $.<\/p>\n
\u00ad<\/p>\n
$\"\"$ <\/a>Comecemos por considerar a corda [AC].<\/p>\n
O \u00e2ngulo CAD \u00e9 um \u00e2ngulo inscrito, cuja amplitude \u00e9 $C\\widehat AD = \\frac{{\\mathop {CD}\\limits^\\frown\u00a0 }}{2} = \\frac{{100^\\circ }}{2} = 50^\\circ $.<\/p>\n
O \u00e2ngulo CAD \u00e9 um \u00e2ngulo externo do tri\u00e2ngulo [ACM], pelo que a sua amplitude \u00e9 igual \u00e0 soma das amplitudes dos \u00e2ngulos internos n\u00e3o adjacentes, isto \u00e9: $C\\widehat AD = A\\widehat CM + A\\widehat MC$.<\/p>\n
Logo, $A\\widehat CM = C\\widehat AD – A\\widehat MC = 50^\\circ\u00a0 – 20^\\circ\u00a0 = 30^\\circ $.<\/p>\n
Como o \u00e2ngulo\u00a0ACB \u00e9 um \u00e2ngulo inscrito, a amplitude do arco compreendido entre os seus lados \u00e9 dupla da amplitude deste \u00e2ngulo inscrito. Assim, $\\mathop {AB}\\limits^\\frown\u00a0\u00a0 = 2 \\times A\\widehat CB = 2 \\times 30^\\circ\u00a0 = 60^\\circ $.<\/p>\n
Logo, $x = 60^\\circ $.<\/p>\n
\u00ad<\/p>\n
$\"\"$ <\/a>Comecemos por considerar a corda [AC].<\/p>\n
Os \u00e2ngulos CAD e ACB s\u00e3o \u00e2ngulos inscritos, pelo que as suas amplitudes s\u00e3o metade das amplitudes dos arcos compreendidos entre os seus lados. Assim, temos:<\/p>\n
$$\\begin{array}{*{20}{c}}
\n{C\\widehat AD = \\frac{{\\mathop {CD}\\limits^\\frown\u00a0 }}{2} = \\frac{{130^\\circ }}{2} = 65^\\circ }&{\\text{e}}&{A\\widehat CB = \\frac{{\\mathop {AB}\\limits^\\frown\u00a0 }}{2} = \\frac{{40^\\circ }}{2} = 20^\\circ }
\n\\end{array}$$<\/p>\n
O \u00e2ngulo CAD \u00e9 um \u00e2ngulo externo do tri\u00e2ngulo [ACM], pelo que a sua amplitude \u00e9 igual \u00e0 soma das amplitudes dos \u00e2ngulos internos n\u00e3o adjacentes, isto \u00e9: $C\\widehat AD = A\\widehat CM + A\\widehat MC$.<\/p>\n
Logo, $A\\widehat MC = C\\widehat AD – A\\widehat CM = 65^\\circ\u00a0 – 20^\\circ\u00a0 = 45^\\circ $.<\/p>\n
Portanto, $x = 45^\\circ $.<\/p>\n
<< Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"
Enunciado Resolu\u00e7\u00e3o Enunciado Observa as figuras e calcula, em cada caso, o valor de x. Resolu\u00e7\u00e3o >> Resolu\u00e7\u00e3o << Enunciado<\/p>\n","protected":false},"author":1,"featured_media":20416,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[213,97,278],"tags":[426,280,188],"series":[],"class_list":["post-7373","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-9--ano","category-aplicando","category-circunferencia-e-poligonos","tag-9-o-ano","tag-angulo-inscrito","tag-circunferencia"],"views":2957,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/9CA-Pag030-3_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7373","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=7373"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7373\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20416"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=7373"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=7373"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=7373"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=7373"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}