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A point on the unit circle is (12,3√2) . What is the sine of the angle formed by a terminal ray that intersects the unit circle...Asked by yeslian c
A point on the unit circle is (12,3√2) . What is the sine of the angle formed by a terminal ray that intersects the unit circle at this point?(1 point)
Responses
3√2
Start Fraction Start Root 3 End Root over 2 End Fraction
−3√2
negative Start Fraction Start Root 3 End Root over 2 End Fraction
−12
negative Start Fraction 1 over 2 End Fraction
12
Start Fraction 1 over 2 End Fraction
Responses
3√2
Start Fraction Start Root 3 End Root over 2 End Fraction
−3√2
negative Start Fraction Start Root 3 End Root over 2 End Fraction
−12
negative Start Fraction 1 over 2 End Fraction
12
Start Fraction 1 over 2 End Fraction
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To determine the sine of the angle formed by the terminal ray that intersects the unit circle at the point \((12, 3\sqrt{2})\), we need to note that this point does not lie on the unit circle since the unit circle is defined by the equation \(x^2 + y^2 = 1\).
Let's check if the point lies on the unit circle by calculating:
\[
x^2 + y^2 = 12^2 + (3\sqrt{2})^2 = 144 + 18 = 162
\]
Since \(162 \neq 1\), the point \((12, 3\sqrt{2})\) is not on the unit circle.
However, if we want to find the sine of the angle corresponding to the ray that intersects the unit circle, we can use the \(y\)-coordinate of the point on the unit circle.
To find the coordinates of the point on the unit circle that corresponds to this ray, we can scale down the coordinates. The distance from the origin to the point \((12, 3\sqrt{2})\) is:
\[
r = \sqrt{12^2 + (3\sqrt{2})^2} = \sqrt{144 + 18} = \sqrt{162} = 9\sqrt{2}
\]
The unit circle has a radius of \(1\). Therefore, we can find the normalized coordinates of the corresponding point on the unit circle:
\[
\left(\frac{12}{9\sqrt{2}}, \frac{3\sqrt{2}}{9\sqrt{2}}\right) = \left(\frac{4}{3\sqrt{2}}, \frac{1}{3}\right)
\]
Now, we focus on the \(y\)-coordinate since:
\[
\sin(\theta) = y \text{ (value on unit circle)}
\]
The \(y\)-coordinate (sine value) of the point on the unit circle is:
\[
\frac{1}{3}
\]
Comparing this with the options given:
- \(3\sqrt{2}\)
- \(\frac{\sqrt{3}}{2}\)
- \(-3\sqrt{2}\)
- \(-\frac{\sqrt{3}}{2}\)
- \(-12\)
- \(\frac{1}{2}\)
None of the provided answers include \(\frac{1}{3}\). Please review the point and recheck it or clarify the context if there's a specific desired outcome or format needed. However, the sine of the angle corresponding to the terminal ray that intersects the unit circle from the original point is \( \frac{1}{3} \).
Let's check if the point lies on the unit circle by calculating:
\[
x^2 + y^2 = 12^2 + (3\sqrt{2})^2 = 144 + 18 = 162
\]
Since \(162 \neq 1\), the point \((12, 3\sqrt{2})\) is not on the unit circle.
However, if we want to find the sine of the angle corresponding to the ray that intersects the unit circle, we can use the \(y\)-coordinate of the point on the unit circle.
To find the coordinates of the point on the unit circle that corresponds to this ray, we can scale down the coordinates. The distance from the origin to the point \((12, 3\sqrt{2})\) is:
\[
r = \sqrt{12^2 + (3\sqrt{2})^2} = \sqrt{144 + 18} = \sqrt{162} = 9\sqrt{2}
\]
The unit circle has a radius of \(1\). Therefore, we can find the normalized coordinates of the corresponding point on the unit circle:
\[
\left(\frac{12}{9\sqrt{2}}, \frac{3\sqrt{2}}{9\sqrt{2}}\right) = \left(\frac{4}{3\sqrt{2}}, \frac{1}{3}\right)
\]
Now, we focus on the \(y\)-coordinate since:
\[
\sin(\theta) = y \text{ (value on unit circle)}
\]
The \(y\)-coordinate (sine value) of the point on the unit circle is:
\[
\frac{1}{3}
\]
Comparing this with the options given:
- \(3\sqrt{2}\)
- \(\frac{\sqrt{3}}{2}\)
- \(-3\sqrt{2}\)
- \(-\frac{\sqrt{3}}{2}\)
- \(-12\)
- \(\frac{1}{2}\)
None of the provided answers include \(\frac{1}{3}\). Please review the point and recheck it or clarify the context if there's a specific desired outcome or format needed. However, the sine of the angle corresponding to the terminal ray that intersects the unit circle from the original point is \( \frac{1}{3} \).
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