Question

Alguém me ajuda com essas conta pra hj por favor​

Answer 1

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

$\Large\displaystyle\boxed{\boxed{\sf x = 30 ^\circ} \: \boxed{\sf y = 150 ^\circ}\: \boxed{z = 150^\circ}}$

No item j, temos que:

$\Large\displaystyle\boxed{\boxed{\sf x = 50 ^\circ} \: \boxed{\sf y = 50 ^\circ}\: \boxed{\sf z =130^\circ}}$

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

No item i)

$\Large\displaystyle{\sf x =30^\circ}\\\\\Large\displaystyle{\sf y = z }$

A soma de todos os ângulos é igual a 360 º

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

Tendo que:

$\Large\displaystyle\boxed{\boxed{\sf x = 30 ^\circ} \: \boxed{\sf y = 150 ^\circ}\: \boxed{z = 150^\circ}}$

No item j)

$\Large\displaystyle{\sf z =x+80^\circ}\\\\\Large\displaystyle{\sf x = y }$

A soma de todos os ângulos é igual a 360 º

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

$\Large\displaystyle{\sf z =x+80}\\\\\Large\displaystyle{\sf z = 50+80 }\\\\\Large\displaystyle{\sf z =130^\circ}$

Tendo que:

$\Large\displaystyle\boxed{\boxed{\sf x = 50 ^\circ} \: \boxed{\sf y = 50 ^\circ}\: \boxed{\sf z =130^\circ}}$

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