Seção II: Ângulos ⇒ Problema 138 - Linhas Retas e Ângulos-Vol. 1 Tópico resolvido

[petrasMOD]

Dados os ángulos adjacentes suplementares AOB e BOC, se traçam
os raios Ox e Oy suas bissetrizes e os raios OP y OQ com a condição :
[tex3]2m\angle AOP = m \angle POy,[/tex3] e [tex3]2m\angle COQ = m\angle QOx[/tex3].
Calcular a medida do ângulo que formam as bissetrizes dos ângulos xOy e POQ.
Se : [tex3]m \angle AOB - m \angle BOC = k^o [/tex3]

Resposta

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Gabarito: D) [tex3]\frac{k^o}{6}[/tex3]

[petrasMOD]

[tex3]m\angle AOB = \alpha, m\angle BOC = \beta.[/tex3]
AOB e BOC são adjacentes suplementares [tex3]\implies \alpha + \beta = 180^\circ.[/tex3]
[tex3]m\angle AOB - m\angle BOC = k^\circ \implies \alpha - \beta = k^\circ.[/tex3]
Resolvendo o sistema: [tex3]\alpha = 90^\circ + \frac{k}{2}~ e~ \beta = 90^\circ - \frac{k}{2}.[/tex3]

OA=[tex3]0^\circ[/tex3]:
[tex3]Ox (bissetriz de AOB):m\angle AO_x = \frac{\alpha}{2}\\.
Oy (bissetriz de BOC): m\angle AO_y = m\angle AOB + m\angle BO_y = \alpha + \frac{\beta}{2}.[/tex3]

[tex3]m\angle AO_y = m\angle AOP + m\angle PO_y = m\angle AOP + 2 \cdot m\angle AOP = 3 \cdot m\angle AOP.\\
m\angle AOP = \frac{1}{3} m\angle AO_y = \frac{1}{3}\left(\alpha + \frac{\beta}{2}\right) = \frac{\alpha}{3} + \frac{\beta}{6}.\\
m\angle CO_x = m\angle COB + m\angle BO_x = \beta + \frac{\alpha}{2}.\\
m\angle CO_x = m\angle COQ + m\angle QO_x = m\angle COQ + 2 \cdot m\angle COQ = 3 \cdot m\angle COQ.\\
m\angle COQ = \frac{1}{3} m\angle CO_x = \frac{1}{3}\left(\beta + \frac{\alpha}{2}\right) = \frac{\beta}{3} + \frac{\alpha}{6}.[/tex3]

Posição de OQ a partir de OA: [tex3]m\angle AOQ = 180^\circ - m\angle COQ = 180^\circ - \left(\frac{\beta}{3} + \frac{\alpha}{6}\right).
[/tex3]

Bissetriz de xOy (OR):
[tex3]m\angle xOy = m\angle AO_y - m\angle AO_x = (\alpha + \frac{\beta}{2}) - \frac{\alpha}{2} = \frac{\alpha + \beta}{2} = \frac{180^\circ}{2} = 90^\circ.[/tex3]
Posição de OR:[tex3]m\angle AOR = m\angle AO_x + \frac{1}{2}m\angle xOy = \frac{\alpha}{2} + \frac{90^\circ}{2} = \frac{\alpha}{2} + 45^\circ.[/tex3]

Bissetriz de POQ (OS):
[tex3]m\angle POQ = m\angle AOQ - m\angle AOP = \left(180^\circ - \frac{\beta}{3} - \frac{\alpha}{6}\right) - \left(\frac{\alpha}{3} + \frac{\beta}{6}\right)\\
m\angle POQ = 180^\circ - \left(\frac{\alpha}{3} + \frac{\alpha}{6}\right) - \left(\frac{\beta}{3} + \frac{\beta}{6}\right) = 180^\circ - \frac{3\alpha}{6} - \frac{3\beta}{6} = 180^\circ - \frac{\alpha+\beta}{2} = 180^\circ - 90^\circ = 90^\circ.
[/tex3]
Posição de OS:[tex3] m\angle AOS = m\angle AOP + \frac{1}{2}m\angle POQ = \left(\frac{\alpha}{3} + \frac{\beta}{6}\right) + \frac{90^\circ}{2} = \frac{\alpha}{3} + \frac{\beta}{6} + 45^\circ.[/tex3]

Ângulo entre OR e OS ([tex3]m\angle ROS[/tex3]):
[tex3]m\angle ROS = |m\angle AOS - m\angle AOR|\\
m\angle ROS = \left| \left(\frac{\alpha}{3} + \frac{\beta}{6} + 45^\circ\right) - \left(\frac{\alpha}{2} + 45^\circ\right) \right|\\
m\angle ROS = \left| \frac{\alpha}{3} + \frac{\beta}{6} - \frac{\alpha}{2} \right|
= \left| \frac{\beta - \alpha}{6}\right| [/tex3]

[tex3]\alpha - \beta = k \therefore \beta - \alpha = -k.\\
m\angle ROS = \left| \frac{-k}{6} \right|[/tex3]

Como a medida do ângulo é positiva, [tex3]\boxed{m\angle ROS = \frac{k}{6}}[/tex3]