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Home » The Fox Cleverly Jumps Over Obstacles—To Prove Not Every Challenge Is Insurmountable! - AMAZONAWS

The Fox Cleverly Jumps Over Obstacles—To Prove Not Every Challenge Is Insurmountable! - AMAZONAWS

📅 March 5, 2026 👤 scraface

Mar 05, 2026

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📌 5Frage: Berechne \(\cos 150^\circ\).

📌 Lösung: Wir wissen, dass \(\cos(180^\circ - \theta) = -\cos \theta\). Somit ist \(\cos 150^\circ = \cos(180^\circ - 30^\circ) = -\cos 30^\circ\). Unter Verwendung des bekannten Wertes \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), finden wir \(\cos 150^\circ = -\frac{\sqrt{3}}{2}\). Daher ist die Antwort \(\boxed{-\frac{\sqrt{3}}{2}}\).

📌 Frage: Ein gleichseitiges Dreieck hat einen Umfang von 24 cm. Wenn jede Seite des Dreiecks um 2 cm vergrößert wird, um wie viele Quadratzentimeter erhöht sich die Fläche?

📌 Lösung: Die Seitenlänge des ursprünglichen gleichseitigen Dreiecks ist \(\frac{24}{3} = 8\) cm. Die Fläche eines gleichseitigen Dreiecks mit Seitenlänge \(s\) ist \(\frac{\sqrt{3}}{4}s^2\). Die ursprüngliche Fläche ist \(\frac{\sqrt{3}}{4} \times 8^2 = 16\sqrt{3}\) cm\(^2\). Nach der letzten Änderung beträgt die Seitenlänge \(8 + 2 = 10\) cm, und die neue Fläche ist \(\frac{\sqrt{3}}{4} \times 10^2 = 25\sqrt{3}\) cm\(^2\). Die Zunahme der Fläche ist \(25\sqrt{3} - 16\sqrt{3} = 9\sqrt{3}\) cm\(^2\). Daher erhöht sich die Fläche um \(\boxed{9\sqrt{3}}\) cm\(^2\).

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