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A student uses the equation \( \tan \theta=\frac{s^{2}}{49} \) to represent the speed, \( s \), in feet per second, of a toy car driving around a circula track having an angle of incline \( \theta \), where \( \sin \theta=\frac{1}{2} \). To solve the problem, the student used the given value of sin \( \theta \) to find the value of \( \tan \theta \) and then substituted the value of \( \tan \theta \) in the equation above to solve for \( s \). What is the approximate value of \( s \), the speed of the car in feet per second? 5.3 7.5 9.2 28.3

Ask by Wagner Bryant. in the United States
Mar 11,2025

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The speed of the car is approximately 5.3 feet per second.

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**Step 1.** Since we are given \(\sin\theta=\frac{1}{2}\) and considering the context of an inclined track (where \(\theta\) is an acute angle), we take \[ \theta=30^\circ. \] **Step 2.** Compute \(\tan\theta\) using the known value of the tangent for \(30^\circ\): \[ \tan30^\circ=\frac{1}{\sqrt{3}}. \] **Step 3.** The given equation is: \[ \tan\theta=\frac{s^2}{49}. \] Substitute \(\tan\theta=\frac{1}{\sqrt{3}}\) into the equation: \[ \frac{1}{\sqrt{3}}=\frac{s^2}{49}. \] **Step 4.** Solve for \(s^2\) by multiplying both sides of the equation by 49: \[ s^2=49\cdot\frac{1}{\sqrt{3}}=\frac{49}{\sqrt{3}}. \] **Step 5.** Take the square root of both sides to solve for \(s\): \[ s=\sqrt{\frac{49}{\sqrt{3}}}. \] Simplify by expressing \(49\) as \(7^2\): \[ s=\frac{7}{\sqrt[4]{3}}. \] **Step 6.** For an approximate numerical value, first compute: \[ \sqrt{3} \approx 1.732, \] thus, \[ \frac{49}{\sqrt{3}} \approx \frac{49}{1.732}\approx 28.28. \] Then, \[ s\approx \sqrt{28.28}\approx 5.32. \] **Step 7.** Rounding to one decimal place: \[ s\approx 5.3 \text{ ft/s}. \] Thus, the speed of the car is approximately \(5.3\) feet per second.

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First, let's determine the value of \( \tan \theta \) given \( \sin \theta = \frac{1}{2} \). Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we find \( \cos \theta = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \). Therefore, \( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \). Substituting this into the equation \( \tan \theta=\frac{s^{2}}{49} \), we have \( \frac{1}{\sqrt{3}}=\frac{s^{2}}{49} \), leading to \( s^2 = \frac{49}{\sqrt{3}} \), and approximating \( s \), we find \( s \approx 28.3 \) feet per second. To double-check our speed, remember that in trigonometry, \( \sin \theta \) gives the opposite side over the hypotenuse while \( \tan \theta \) is related to the angle's slope, helping you understand how steep your incline is. In practical terms, knowing how to balance these relationships can help in real-life applications like designing racetracks or calculating vehicle dynamics on inclines!

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