- Racso.png (32.02 KiB) Exibido 468 vezes
Seja [tex3]A\hat{O}B = 2\alpha[/tex3]
e [tex3]B\hat{O}C = 2\theta[/tex3]
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Então, [tex3]A\hat{O}P = P\hat{O}B = \alpha[/tex3]
e [tex3]B\hat{O}Q= C\hat{O}Q =\theta[/tex3]
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Como [tex3]2\alpha + 2\theta = 300° \implies \alpha+ \theta = 150°[/tex3]
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[tex3]A\hat{O}R = R\hat{O}Q = \dfrac{2\alpha+ \theta}{2} = \alpha + \dfrac{\theta}{2}[/tex3]
Como [tex3]R\hat{O}Q = R\hat{O}B+B\hat{O}Q \implies \alpha+\dfrac{\theta}{2} = R\hat{O}B + \theta \implies R\hat{O}B = \alpha - \dfrac{\theta}{2}[/tex3]
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Analogamente:
[tex3]P\hat{O}S = C\hat{O}S = \dfrac{2\theta + \alpha}{2} = \theta + \dfrac{\alpha}{2}[/tex3]
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Como [tex3]P\hat{O}S = P\hat{O}B+ B\hat{O}S \implies \theta + \dfrac{\alpha}{2} = \alpha+B\hat{O}S \implies B\hat{O}S = \theta - \dfrac{\alpha}{2}[/tex3]
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[tex3]R\hat{O}S = R\hat{O}B+B\hat{O}S = \alpha - \dfrac{\theta}{2}+ \theta - \dfrac{\alpha}{2} = \dfrac{\alpha}{2} + \dfrac{\theta}{2} = \boxed{75°}[/tex3]