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)
$ heta$
$=\frac{\pi}{2}, r$ $\qquad$
9

Q: 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

∻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. ⚹◊What is the formula to calculate the coordinates of a point on a circle given the angle $\theta$ and radius $r$?⍭ Generate me a similar question◊

Question

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)
$	heta$
$=\frac{\pi}{2}, r$ $\qquad$
9

Accepted Answer

∻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. ⚹◊What is the formula to calculate the coordinates of a point on a circle given the angle $\theta$ and radius $r$?⍭ Generate me a similar question◊

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