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À primeira vista parece um problema complicado de geometria, mas ele se resolve rapidamente observando ângulos suplementares e a soma dos ângulos internos de um triângulo. No topo da figura existe uma reta e dois ângulos de 50° formados com os lados do triângulo. Esses ângulos estão fora do triângulo. O ângulo interno do vértice superior é suplementar a esses dois ângulos. Ângulos suplementares somam 180°. Assim, o ângulo interno do vértice do triângulo é: 180° − 50° − 50° = 80° Agora usamo
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Desafio geométrico que parece simples, mas exige atenção aos detalhes. O ângulo externo indicado mede 110°. Como ele forma um par linear com o ângulo interno da base do triângulo, calculamos o seu suplementar. Ângulos suplementares somam 180°, portanto 180° − 110° = 70°. Esse ângulo de 70° é um dos ângulos internos do triângulo. Observando as marcações, percebemos que o triângulo é isósceles, pois possui dois lados congruentes. Em um triângulo isósceles, os ângulos da base são congruentes. Log
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O ângulo reto interno é bom, rápido e fácil de fazê-lo.
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