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Juan Burgos

Juan Burgos. 6e. Punto # 14 Octubre. FC ┴. EA. <FAB=?. D. E. 40°. C. 30°. B. F. A. FC ┴. EA. m <FAB =?. D. E. 40°. C. 30°. 90°. 90°. B. Como los segmento s son perpendiculares, sus ángulos miden 90º. 90°. F. A. FC ┴. EA. m<FAB =?. D. E. 40°. C. 30°. B.

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Juan Burgos

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  1. Juan Burgos 6e Punto # 14 Octubre

  2. FC ┴ EA <FAB=? D E 40° C 30° B F A

  3. FC ┴ EA m<FAB=? D E 40° C 30° 90° 90° B Como los segmento s son perpendiculares, sus ángulos miden 90º 90° F A

  4. FC ┴ EA m<FAB=? D E 40° C 30° B 110° El àngulo DFC es igual a 110º por suma de ángulos del triangulo DFC 90° F A

  5. FC ┴ EA m<FAB=? D E 40° C 30° El ángulo BFA mide 70º por ser suplementario con el ángulo DFB, que mide 110º B 110° 90° 70° F A

  6. FC ┴ EA m<FAB=20º D E 40° C 30 RESPUESTA El ángulo FAB es igual a 20º por suma interna de angulos de triangulo FAB B 110° 90° 70°| F 20° A

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