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Home » From Beginners to Experts: Perfect ‘G’ in Cursive Fast—Watch Now! - Crankk.io

From Beginners to Experts: Perfect ‘G’ in Cursive Fast—Watch Now! - Crankk.io

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 Um \(\tan 75^\circ\) zu berechnen, verwenden wir die Additionstheorem für Tangens:

📌 \tan(75^\circ) = \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ}.

📌 Wir wissen, dass \(\tan 45^\circ = 1\) und \(\tan 30^\circ = \frac{1}{\sqrt{3}}\). Einsetzen:

📌 \tan 75^\circ = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}.

📌 Rationalisieren des Nenners:

📌 \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - (1)^2} = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}.

📌 Somit ist \(\tan 75^\circ = \boxed{2 + \sqrt{3}}\).

📌 Frage: Ein gleichschenkliges Dreieck hat eine Basis von 10 cm und neue Höhen, die von den Basisecken zum gegenüberliegenden Scheitelpunkt gezogen werden, wodurch es in drei kleinere Dreiecke unterteilt wird. Wenn jede der neuen Dreiecke eine Fläche von 15 cm² hat, wie lang ist die Höhe des ursprünglichen gleichschenkligen Dreiecks?

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