✌Rpta: El valor de "x" es 39
[tex]\boldsymbol {\mathsf {Procedimiento}}[/tex]
Resolvemos por el toerema de los triangulos y es : La suma de dos angulos internos es igual a la suma de un angulo externo
[tex]\begin {gathered}{\mathsf {x + 41^{o} = 80^{o}}}\end{gathered}[/tex]
[tex]\begin {gathered}{\mathsf {x = 80^{o} - 41^{o}}}\end{gathered}[/tex]
[tex]\begin {gathered}{\mathsf {x = 39^{o}}}\end{gathered}[/tex]
Atentamente : [tex]\begin {gathered}\mathsf{\boxed{\bold {E}}_{\boxed {\bold {N}}}}\end {gathered}[/tex][tex]\begin {gathered}\mathsf{\boxed{\bold {V}}_{\boxed {\bold {E}}}}\end {gathered}[/tex][tex]\mathsf{\boxed {\bold{R}}}[/tex]
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✌Rpta: El valor de "x" es 39
[tex]\boldsymbol {\mathsf {Procedimiento}}[/tex]
Resolvemos por el toerema de los triangulos y es : La suma de dos angulos internos es igual a la suma de un angulo externo
[tex]\begin {gathered}{\mathsf {x + 41^{o} = 80^{o}}}\end{gathered}[/tex]
[tex]\begin {gathered}{\mathsf {x = 80^{o} - 41^{o}}}\end{gathered}[/tex]
[tex]\begin {gathered}{\mathsf {x = 39^{o}}}\end{gathered}[/tex]
Atentamente : [tex]\begin {gathered}\mathsf{\boxed{\bold {E}}_{\boxed {\bold {N}}}}\end {gathered}[/tex][tex]\begin {gathered}\mathsf{\boxed{\bold {V}}_{\boxed {\bold {E}}}}\end {gathered}[/tex][tex]\mathsf{\boxed {\bold{R}}}[/tex]