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do teachers get paid in the summer - Deep Underground Poetry

📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 \Delta A = A_2 - A_1 = 36\sqrt{3} - 25\sqrt{3} = 11\sqrt{3}

📌 Thus, the area increases by \(\boxed{11\sqrt{3}}\) square centimeters.

📌 Compute \(\cos 60^\circ\) using the unit circle and verify using the properties of an equilateral triangle.

📌 On the unit circle, the coordinates of the point corresponding to \(60^\circ\) are \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\). Thus, \(\cos 60^\circ = \frac{1}{2}\).

📌 Alternatively, consider an equilateral triangle with side length 2. By splitting it into two 30-60-90 right triangles, the side opposite \(60^\circ\) is \(\sqrt{3}\), and the hypotenuse is 2. Therefore:

📌 \cos 60^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{2}

📌 Thus, \(\cos 60^\circ\) is \(\boxed{\frac{1}{2}}\).Question: Find $\tan \theta$ where $\theta$ is the angle between the line of sight from a weather balloon at $(2, 3)$ to a cloud at $(5, 11)$ and the horizontal axis.

📌 Solution: The slope of the line from $(2, 3)$ to $(5, 11)$ is $\frac{11 - 3}{5 - 2} = \frac{8}{3}$. Since $\tan \theta$ equals the slope, $\tan \theta = \frac{8}{3}$. \boxed{\dfrac{8}{3}}

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