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Sagot :
To find the speed [tex]\( s \)[/tex] of the toy car, follow these steps:
1. Given information and the relevant relationship:
- [tex]\( \sin \theta = \frac{1}{2} \)[/tex]
- The equation [tex]\( \tan \theta = \frac{s^2}{49} \)[/tex] represents the relationship between the speed [tex]\( s \)[/tex] and the angle [tex]\( \theta \)[/tex].
2. Find [tex]\( \theta \)[/tex]:
Since [tex]\( \sin \theta = \frac{1}{2} \)[/tex], we identify [tex]\( \theta \)[/tex] corresponding to this sine value.
- The angle [tex]\( \theta \)[/tex] whose sine is [tex]\( \frac{1}{2} \)[/tex] is [tex]\( \frac{\pi}{6} \)[/tex] radians, or [tex]\( 30^\circ \)[/tex].
3. Calculate [tex]\( \tan \theta \)[/tex]:
Using the relationship of trigonometric functions, for [tex]\( \theta = \frac{\pi}{6} \)[/tex]:
[tex]\[ \tan \left( \frac{\pi}{6} \right) = \frac{\sin \left( \frac{\pi}{6} \right)}{\cos \left( \frac{\pi}{6} \right)} \][/tex]
Given [tex]\( \sin \theta = \frac{1}{2} \)[/tex] and knowing [tex]\( \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} \)[/tex]:
[tex]\[ \tan \left( \frac{\pi}{6} \right) = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \approx 0.577 \][/tex]
4. Solve for [tex]\( s \)[/tex]:
The given equation is [tex]\( \tan \theta = \frac{s^2}{49} \)[/tex]. Substitute [tex]\( \tan \theta \)[/tex] into this equation:
[tex]\[ 0.577 = \frac{s^2}{49} \][/tex]
Solve for [tex]\( s^2 \)[/tex]:
[tex]\[ s^2 = 0.577 \times 49 \][/tex]
[tex]\[ s^2 \approx 28.273 \][/tex]
Take the square root of both sides to find [tex]\( s \)[/tex]:
[tex]\[ s \approx \sqrt{28.273} \approx 5.318 \][/tex]
5. Interpret the result in the context of provided choices:
Among the given choices, the approximate value of [tex]\( s \)[/tex] is:
[tex]\[ \boxed{5.3} \][/tex]
Therefore, the speed [tex]\( s \)[/tex] of the car in feet per second is approximately 5.3 feet per second.
1. Given information and the relevant relationship:
- [tex]\( \sin \theta = \frac{1}{2} \)[/tex]
- The equation [tex]\( \tan \theta = \frac{s^2}{49} \)[/tex] represents the relationship between the speed [tex]\( s \)[/tex] and the angle [tex]\( \theta \)[/tex].
2. Find [tex]\( \theta \)[/tex]:
Since [tex]\( \sin \theta = \frac{1}{2} \)[/tex], we identify [tex]\( \theta \)[/tex] corresponding to this sine value.
- The angle [tex]\( \theta \)[/tex] whose sine is [tex]\( \frac{1}{2} \)[/tex] is [tex]\( \frac{\pi}{6} \)[/tex] radians, or [tex]\( 30^\circ \)[/tex].
3. Calculate [tex]\( \tan \theta \)[/tex]:
Using the relationship of trigonometric functions, for [tex]\( \theta = \frac{\pi}{6} \)[/tex]:
[tex]\[ \tan \left( \frac{\pi}{6} \right) = \frac{\sin \left( \frac{\pi}{6} \right)}{\cos \left( \frac{\pi}{6} \right)} \][/tex]
Given [tex]\( \sin \theta = \frac{1}{2} \)[/tex] and knowing [tex]\( \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} \)[/tex]:
[tex]\[ \tan \left( \frac{\pi}{6} \right) = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \approx 0.577 \][/tex]
4. Solve for [tex]\( s \)[/tex]:
The given equation is [tex]\( \tan \theta = \frac{s^2}{49} \)[/tex]. Substitute [tex]\( \tan \theta \)[/tex] into this equation:
[tex]\[ 0.577 = \frac{s^2}{49} \][/tex]
Solve for [tex]\( s^2 \)[/tex]:
[tex]\[ s^2 = 0.577 \times 49 \][/tex]
[tex]\[ s^2 \approx 28.273 \][/tex]
Take the square root of both sides to find [tex]\( s \)[/tex]:
[tex]\[ s \approx \sqrt{28.273} \approx 5.318 \][/tex]
5. Interpret the result in the context of provided choices:
Among the given choices, the approximate value of [tex]\( s \)[/tex] is:
[tex]\[ \boxed{5.3} \][/tex]
Therefore, the speed [tex]\( s \)[/tex] of the car in feet per second is approximately 5.3 feet per second.
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