How do you write in simplest radical form the coordinates of point A if A is on the terminal side of angle in standard position whose degree measure is $θ$ $theta$ : $O A = \sqrt{3}$ $OA=sqrt3$ , $θ = 300^{\circ}$ $theta=300^circ$ ?

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Answer 1 · 2018-03-04T07:18:54

Coordinates of $A (\frac{\sqrt{3}}{2}, - \frac{3}{2})$ $A (sqrt3/2, -3/2)$

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Given $θ = 300^{\circ}, O A = = \sqrt{3}$ $theta = 300^@, OA ==sqrt 3$

To find the coordinates of $A (x, y)$ $A(x,y)$

$300^{\circ}$ $300^@$ is in fourth quadrant.

OA forms a triangle with x axis and the three angles are having the measurements of $30^{\circ}, 60^{\circ}, 90^{\circ}$ $30^@, 60^@, 90^@$ .

Then the sides will be in the ratio $x : \sqrt{3} x = 2 x$ $x : sqrt 3 x = 2x$

$x = ¯ ¯¯¯¯ ¯ O A cos θ = \sqrt{3} \cdot cos 300 = \sqrt{3} cos - 60$ $x = bar(OA) cos theta = sqrt 3 * cos 300 = sqrt 3 cos -60$

$x = \sqrt{3} cos 60 = \frac{\sqrt{3} \cdot \sqrt{1}}{2} = \frac{\sqrt{3}}{2}$ $x = sqrt 3 cos 60 = (sqrt 3 * sqrt 1) / 2 = sqrt3 /2$

$y = ¯ ¯¯¯¯ ¯ O A sin θ = \sqrt{3} sin 300 = - \sqrt{3} sin 60$ $y = bar(OA) sin theta = sqrt 3 sin 300 = - sqrt 3 sin 60$

$y = - \frac{\sqrt{3} \cdot \sqrt{3}}{2} = - \frac{3}{2}$ $y = - (sqrt 3 * sqrt 3) / 2 = -3/2$

How do you write in simplest radical form the coordinates of point A if A is on the terminal side of angle in standard position whose degree measure is θtheta: OA=√3OA=sqrt3, θ=300∘theta=300^circ?