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An angle in standard position in the $\boldsymbol{x} \boldsymbol{y}$-plane is given in radians. A circle is centered at the origin with the given radius.
What are the coordinates of the point of intersection of the terminal ray of the angle and the circle? *
[4) (1分)
$\theta$
$=\frac{\pi}{2}, r$ $\qquad$
9
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Answer from Sia

Posted about 2 months ago

Solution by Steps

step 1

To find the radius $r$, we use the distance formula from the origin to the point $(-8\sqrt{2}, -8\sqrt{2})$: $r = \sqrt{(-8\sqrt{2})^2 + (-8\sqrt{2})^2}$

step 2

Calculating $r$: $r = \sqrt{128 + 128} = \sqrt{256} = 16$

step 3

To find the angle $\theta$, we use the tangent function: $\tan(\theta) = \frac{y}{x} = \frac{-8\sqrt{2}}{-8\sqrt{2}} = 1$

step 4

Since the point $(-8\sqrt{2}, -8\sqrt{2})$ is in the third quadrant, we take $\theta = \frac{5\pi}{4}$

Answer

$r = 16$ and $\theta = \frac{5\pi}{4}$

Key Concept

The relationship between the coordinates of a point and the radius and angle in polar coordinates.

Explanation

The radius is determined by the distance from the origin to the point, and the angle is found using the tangent function, adjusted for the quadrant.

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