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Home » t = -\frac-122 \times 3 = \frac126 = 2 - Baxtercollege

t = -\frac-122 \times 3 = \frac126 = 2 - Baxtercollege

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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📌 Thus, the number of distinct real roots is \(\boxed{3}\).Question: Compute $\tan 30^\circ$ in a right triangle with an opposite side of length $1$ and adjacent side of length $\sqrt{3}$.

📌 Solution: By definition, $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$. Substituting the given values, $\tan 30^\circ = \frac{1}{\sqrt{3}}$. Rationalizing the denominator, $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$. Thus, $\boxed{\dfrac{\sqrt{3}}{3}}$.

📌 Question: A sundial's gnomon casts a shadow of length $5\,\text{cm}$ when the sun's elevation angle is $60^\circ$. What is the height of the gnomon?

📌 Solution: Using $\tan(60^\circ) = \frac{\text{height}}{\text{shadow length}}$, we substitute $\tan(60^\circ) = \sqrt{3}$ and shadow length $5\,\text{cm}$. Solving for height: $\text{height} = 5 \cdot \sqrt{3}$. Thus, $\boxed{5\sqrt{3}}$.

📌 Question: An equilateral triangle has an area of $25\sqrt{3}\,\text{cm}^2$. If each side is increased by $4\,\text{cm}$, by how many square centimeters does the area increase?

📌 Solution: The area of an equilateral triangle is $\frac{\sqrt{3}}{4} s^2$. Solving for original side $s$: $\frac{\sqrt{3}}{4} s^2 = 25\sqrt{3} \Rightarrow s^2 = 100 \Rightarrow s = 10\,\text{cm}$. New side: $14\,\text{cm}$. New area: $\frac{\sqrt{3}}{4} \cdot 14^2 = 49\sqrt{3}$. Increase: $49\sqrt{3} - 25\sqrt{3} = 24\sqrt{3}$. Thus, $\boxed{24\sqrt{3}}$.

📌 Question: A historical astrolabe features a right triangle

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