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Home » Similarly, $ z^3 = \omega^2 = e^-2\pi i / 3 $, so roots: - Ready Digital AB

Similarly, $ z^3 = \omega^2 = e^-2\pi i / 3 $, so roots: - Ready Digital AB

📅 March 7, 2026 👤 scraface

Mar 07, 2026

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A_{\text{original}} = \frac{1}{2} \cdot 9 \cdot 12 = 54 \text{ cm}^2
The area of the equilateral triangle is:
A_{\text{new}} = \frac{\sqrt{3}}{4} \cdot 12^2 = \frac{\sqrt{3}}{4} \c...
The increase in area is:
\Delta A = 36\sqrt{3} - 54
\boxed{36\sqrt{3} - 54}
Question: Compute $\tan 75^\circ$ using angle addition identities.
Solution: Using the identity:
\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}
Let $a = 45^\circ$, $b = 30^\circ$, so:

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