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A student uses the equation [tex]\tan \theta=\frac{s^2}{49}[/tex] to represent the speed, [tex]s[/tex], in feet per second, of a toy car driving around a circular track having an angle of incline [tex]\theta[/tex], where [tex]\sin \theta=\frac{1}{2}[/tex]. To solve the problem, the student used the given value of [tex]\sin \theta[/tex] to find the value of [tex]\tan \theta[/tex] and then substituted the value of [tex]\tan \theta[/tex] in the equation above to solve for [tex]s[/tex].

What is the approximate value of [tex]s[/tex], the speed of the car in feet per second?

A. 5.3
B. 7.5
C. 9.2
D. 28.3



Answer :

GinnyAnswer
Let's take a detailed step-by-step approach to solve this problem:

1. Given Information:
- [tex]\(\sin \theta = \frac{1}{2}\)[/tex]
- The equation [tex]\(\tan \theta = \frac{s^2}{49}\)[/tex]

2. Find [tex]\(\cos \theta\)[/tex]:
We know the identity involving sine and cosine:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Substituting the given value of [tex]\(\sin \theta = \frac{1}{2}\)[/tex]:
[tex]\[ \left(\frac{1}{2}\right)^2 + \cos^2 \theta = 1 \][/tex]
[tex]\[ \frac{1}{4} + \cos^2 \theta = 1 \][/tex]
[tex]\[ \cos^2 \theta = 1 - \frac{1}{4} \][/tex]
[tex]\[ \cos^2 \theta = \frac{3}{4} \][/tex]
Taking the positive square root (since [tex]\(\cos\)[/tex] is positive in the first and fourth quadrants where sine is also positive):
[tex]\[ \cos \theta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \approx 0.866 \][/tex]

3. Find [tex]\(\tan \theta\)[/tex]:
We use the identity for tangent:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Substituting the values of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \approx 0.577 \][/tex]

4. Substitute [tex]\(\tan \theta\)[/tex] into the equation to find [tex]\(s\)[/tex]:
The given equation is:
[tex]\[ \tan \theta = \frac{s^2}{49} \][/tex]
Substituting the value of [tex]\(\tan \theta\)[/tex]:
[tex]\[ 0.577 \approx \frac{s^2}{49} \][/tex]
Solving for [tex]\(s^2\)[/tex]:
[tex]\[ s^2 = 0.577 \times 49 = 28.3 \][/tex]
Then, taking the square root of both sides gives:
[tex]\[ s = \sqrt{28.3} \approx 5.318 \][/tex]

5. Final Approximate Value:
The speed [tex]\(s\)[/tex] of the car in feet per second is approximately:
[tex]\[ s \approx 5.3 \][/tex]

So, the correct answer from the given options is:
[tex]\[ \boxed{5.3} \][/tex]

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