Verified answer

A partir dos cálculos e o conhecimento sobre ângulos opostos pelo vértice, tem-se que no item i:

[tex]\Large\displaystyle\boxed{\boxed{\sf x = 30 ^\circ} \: \boxed{\sf y = 150 ^\circ}\: \boxed{z = 150^\circ}}[/tex]

No item j, temos que:

[tex]\Large\displaystyle\boxed{\boxed{\sf x = 50 ^\circ} \: \boxed{\sf y = 50 ^\circ}\: \boxed{\sf z =130^\circ}}[/tex]

Os ângulos opostos pelo vértice possuem o mesmo valor (medida).

No item i)

[tex]\Large\displaystyle\text{${\sf x =30^\circ}$}\\\\\Large\displaystyle\text{${\sf y = z }$}[/tex]

A soma de todos os ângulos é igual a 360 º

[tex]\Large\displaystyle\text{${\sf 30+30 +y+y=360}$}\\\\\Large\displaystyle\text{${\sf 60 +2\cdot y=360}$}\\\\\Large\displaystyle\text{${\sf 2\cdot y=360-60}$}\\\\\Large\displaystyle\text{${\sf 2\cdot y=300}$}\\\\\Large\displaystyle\text{${\sf y=\dfrac{300}{2}}$}\\\\\Large\displaystyle\text{${\sf y=150 ^\circ }$}[/tex]

Tendo que:

[tex]\Large\displaystyle\boxed{\boxed{\sf x = 30 ^\circ} \: \boxed{\sf y = 150 ^\circ}\: \boxed{z = 150^\circ}}[/tex]

No item j)

[tex]\Large\displaystyle\text{${\sf z =x+80^\circ}$}\\\\\Large\displaystyle\text{${\sf x = y }$}[/tex]

A soma de todos os ângulos é igual a 360 º

[tex]\Large\displaystyle\text{${\sf x+y+z+x+80=360}$}\\\\\Large\displaystyle\text{${\sf x+x+x+80+x+80=360}$}\\\\\Large\displaystyle\text{${\sf 4 \cdot x+160=360}$}\\\\\Large\displaystyle\text{${\sf 4\cdot x=360-160}$}\\\\\Large\displaystyle\text{${\sf 4\cdot x=200}$}\\\\\Large\displaystyle\text{${\sf x=\dfrac{200}{4}}$}\\\\\Large\displaystyle\text{${\sf x=50 ^\circ }$}[/tex]

[tex]\Large\displaystyle\text{${\sf z =x+80}$}\\\\\Large\displaystyle\text{${\sf z = 50+80 }$}\\\\\Large\displaystyle\text{${\sf z =130^\circ}$}[/tex]

Tendo que:

[tex]\Large\displaystyle\boxed{\boxed{\sf x = 50 ^\circ} \: \boxed{\sf y = 50 ^\circ}\: \boxed{\sf z =130^\circ}}[/tex]

Saiba mais:

brainly.com.br/tarefa/1411497

brainly.com.br/tarefa/742682