Respuesta :

Answer:

[tex]\begin{gathered} x^{\circ}=70^{\circ} \\ y^{\circ}=55^{\circ} \\ z^{\circ}=55^{\circ} \end{gathered}[/tex]

Explanation:

Given that;

[tex]AC=AB[/tex]

then triangle ABC is an isosceles triangle;

[tex]\measuredangle C=\measuredangle B=\frac{180-\measuredangle A}{2}=\frac{180-70}{2}=55^{\circ}[/tex]

From the given figure, we can observe that quadrilateral ADFE is a parallelogram.

Recall that opposite angles of a parallelogram are equal.

So, we have;

[tex]\begin{gathered} x^{\circ}=\measuredangle A \\ x^{\circ}=70^{\circ} \end{gathered}[/tex]

Also, the same rule applies to parallelogram BFDE and CDEF;

So, we have;

[tex]\begin{gathered} y^{\circ}=\measuredangle B=55^{\circ} \\ y^{\circ}=55^{\circ} \end{gathered}[/tex][tex]\begin{gathered} z^{\circ}=\measuredangle C=55^{\circ} \\ z^{\circ}=55^{\circ} \end{gathered}[/tex]

Therefore, we have;

[tex]\begin{gathered} x^{\circ}=70^{\circ} \\ y^{\circ}=55^{\circ} \\ z^{\circ}=55^{\circ} \end{gathered}[/tex]

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