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los ángulos internos de un triangulo suman 180°
A+B+C = 180°
(4x+37)+(5y+67)+(7x-49) = 180
4x+7x+5y+37+67-49 = 180
11x+5y+55 = 180
11x+5y = 180-55
11x+5y = 125 ec. 1
Los ángulos suplementarios suman 180°
(5y+67) + (3x+68) = 180
3x+5y+67+68 = 180
3x+5y=180-67-68
3x+5y = 45 ec. 2
Al multiplicar por -1 a la ec. 2 queda así:
-3x-5y = -45 ec. 3
Al sumar las ec. 1 y 3 queda así:
11x +5y = 125
-3x -5y = -45
⇒ 8x =80
x = 10
De la ec. 2:
3x+ 5y = 45
(3*10) + 5y = 45
30+5y = 45
5y = 45-30
5y = 15
y = 15/5
y = 3
Comprobémoslo en la ec. 1:
11x + 5y = 125
(11*10) + (5*3) = 125
110 + 15 = 125 comprobado
∡E es suplementario del ∡C:
∡E + ∡C = 180
∡C = 7x-49
E+7X-49 = 180
E+(7*10)-49 = 180
E+70-49 = 180
E+21 = 180
E = 180-21
E = 159
Respuesta:
x = 10
y = 3
E = 159°
Saludos.............