Matemática
rafanandafernan
8

O dobro do suplementar do complementar da metade de um ângulo vale 260°. Quanto mede esse ângulo?

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jefersonpartelli

[latex]2\cdot\left[180^\circ-\left(90^\circ-\dfrac{\alpha}{2}\right)\right]=260^\circ\\\\\\\left[180^\circ-\left(90^\circ-\dfrac{\alpha}{2}\right)\right]=130^\circ\\\\\\-\left(90^\circ-\dfrac{\alpha}{2}\right)\right=-50^\circ\\\\\\-90+\dfrac{\alpha}{2}=-50^\circ\\\\\\\dfrac{\alpha}{2}=40^\circ\\\\\boxed{\alpha=80^\circ}[/latex] Esse ângulo mede 80°.

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