Seção II: Ângulos ⇒ Problema 106 - Linhas Retas e Ângulos-Vol. 1 Tópico resolvido

[petrasMOD]

Se OP é bissetriz do [tex3]\angle AOB[/tex3]
Se OQ é bissetriz do [tex3]\angle BOC[/tex3]
Se OM é bissetriz do [tex3]\angle POQ[/tex3]
Calcular [tex3]\angle BOM[/tex3].
[tex3]m\angle BOC - m\angle AOB = 40^o [/tex3]

Resposta

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Gabarito: A) 10^o

[petrasMOD]

[tex3] m\angle POB = \frac{m\angle AOB}{2} (OP :bissetriz de AOB)\\
m\angle BOQ = \frac{m\angle BOC}{2} (OQ: bissetriz de BOC)\\
m\angle POM = m\angle MOQ = \frac{m\angle POQ}{2} (OM :bissetriz de POQ)[/tex3]

[tex3]m\angle POQ = m\angle POB + m\angle BOQ\\
\angle POQ = \frac{m\angle AOB}{2} + \frac{m\angle BOC}{2} = \frac{m\angle AOB + m\angle BOC}{2}[/tex3]

[tex3]m\angle MOQ = \frac{m\angle POQ}{2} = \frac{m\angle AOB + m\angle BOC}{4}
[/tex3]

[tex3]m\angle BOQ = m\angle BOM + m\angle MOQ \implies m\angle BOM = m\angle BOQ - m\angle MOQ[/tex3]
[tex3]m\angle BOM = \frac{m\angle BOC}{2} - \frac{m\angle AOB + m\angle BOC}{4}\\
\therefore m\angle BOM = \frac{m\angle BOC - m\angle AOB}{4}
[/tex3]

[tex3]m\angle BOC - m\angle AOB = 40^\circ\\.
m\angle BOM = \frac{40^\circ}{4}\\
\therefore \boxed{m\angle BOM = 10^\circ}[/tex3]