a student uses the equation $\tan\theta=\frac...
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 circular 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?\n5.3\n7.5\n9.2\n28.3
Answer
# Explanation:
## Step1: Find cosθ
Given $\sin\theta=\frac{1}{2}$, and using $\sin^{2}\theta+\cos^{2}\theta = 1$. So $\cos\theta=\sqrt{1 - \sin^{2}\theta}=\sqrt{1-(\frac{1}{2})^{2}}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$.
## Step2: Calculate tanθ
Since $\tan\theta=\frac{\sin\theta}{\cos\theta}$, substituting $\sin\theta=\frac{1}{2}$ and $\cos\theta=\frac{\sqrt{3}}{2}$, we get $\tan\theta=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}$.
## Step3: Solve for s
We have the equation $\tan\theta=\frac{s^{2}}{49}$, substituting $\tan\theta=\frac{1}{\sqrt{3}}$ into it, we get $\frac{1}{\sqrt{3}}=\frac{s^{2}}{49}$. Then $s^{2}=\frac{49}{\sqrt{3}}\approx\frac{49}{1.732}\approx28.3$. So $s=\sqrt{28.3}\approx 5.3$.
# Answer:
5.3