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Home » \cos 60^\circ = \frac{\textadjacent}{\texthypotenuse} = \frac12 - Aurero

\cos 60^\circ = \frac{\textadjacent}{\texthypotenuse} = \frac12 - Aurero

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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2\sin z + \sqrt{3} = 0 \implies \sin z = -\frac{\sqrt{3}}{2}
الزوايا $z$ التي تحقق $\sin z = -\frac{\sqrt{3}}{2}$ تقع في الربعين ال...
z = 180^\circ + 60^\circ = 240^\circ \quad \text{و} \quad z = 360^\cir...
\boxed{240^\circ} \quad \text{و} \quad \boxed{300^\circ}
الإجابة النهائية:** الزوايا هي $\boxed{240^\circ}$ و $\boxed{300^\circ...
السؤال: أوجد المتجه $\mathbf{v}$ بحيث أن $\mathbf{v} \times \begin{pma...
الحل: ليكن $\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatri...
\mathbf{v} \times \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{p...
نحتاج إلى أنه يساوي $\begin{pmatrix} 0 \\ 3 \\ -2 \end{pmatrix}$, مما...
= 0 \\

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