Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Connect with a community of experts ready to provide precise solutions to your questions on our user-friendly Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
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]
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]
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.