\nEnunciado<\/b><\/span><\/p>\n
Sabendo que $\\hat{B}=62{}^\\text{o}$, $\\overline{AB}=\\overline{BC}$ e $\\overline{CD}=\\overline{CE}$ , calcula $C\\hat{D}E$.<\/p>\n
$\"\"$ <\/p>\n<\/p>\n
Resolu\u00e7\u00e3o >><\/a><\/span><\/div><\/div>\n\n
\nResolu\u00e7\u00e3o<\/b><\/span><\/p>\n
$\"\"$ <\/a>Como sabemos, num tri\u00e2ngulo, a lados geometricamente iguais, op\u00f5em-se \u00e2ngulos geometricamente iguais.<\/p>\n
Logo, no tri\u00e2ngulo [ABC], s\u00e3o geometricamente iguais os \u00e2ngulos BAC e BCA, pois op\u00f5em-se a lados geometricamente iguais.<\/p>\n
Por outro lado, sabemos que a soma das amplitudes dos \u00e2ngulos internos de um tri\u00e2ngulo \u00e9 180\u00ba.<\/p>\n
Logo, temos:<\/p>\n
$B\\hat{A}C=B\\hat{C}A=\\frac{180{}^\\text{o}-A\\hat{B}C}{2}=\\frac{180{}^\\text{o}-62{}^\\text{o}}{2}=59{}^\\text{o}$<\/p>\n
Como os \u00e2ngulos ACB e DCE s\u00e3o verticalmente opostos, ent\u00e3o s\u00e3o geometricamente iguais. Logo, $D\\hat{C}E=A\\hat{C}B=59{}^\\text{o}$.<\/p>\n
Mas, tamb\u00e9m no tri\u00e2ngulo [CDE] h\u00e1 dois lados geometricamente iguais. Pelo que, raciocinando como acima, temos:<\/p>\n
$C\\hat{D}E=C\\hat{E}D=\\frac{180{}^\\text{o}-D\\hat{C}E}{2}=\\frac{180{}^\\text{o}-59{}^\\text{o}}{2}=60,5{}^\\text{o}$<\/p>\n
<< Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"
Enunciado Resolu\u00e7\u00e3o Enunciado Sabendo que $\\hat{B}=62{}^\\text{o}$, $\\overline{AB}=\\overline{BC}$ e $\\overline{CD}=\\overline{CE}$ , calcula $C\\hat{D}E$. Resolu\u00e7\u00e3o >> Resolu\u00e7\u00e3o << Enunciado<\/p>\n","protected":false},"author":1,"featured_media":20630,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,97,102],"tags":[105,67,106],"series":[],"class_list":["post-4617","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-do-espaco-ao-plano","tag-angulos","tag-geometria","tag-triangulos"],"views":2035,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/7V2Pag103-5_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/4617","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=4617"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/4617\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20630"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4617"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=4617"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=4617"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=4617"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}