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 $θ$ $theta$ : OA=6, $θ = 90^{\circ}$ $theta=90^circ$ ?

Question

2460 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-06T13:18:12

Coordinates of $A = (0, 1)$ $color(green)(A = (0,1)$

enter image source here

Point A is in first quadrant as $θ$ $theta$ is $90^{\circ}$ $90^@$

Hence, both x , y coordinates are positive..

$A_{x} = ¯ ¯¯¯¯ ¯ O A cos θ = 6 \cdot cos 90 = 25 \cdot 0 = 0$ $A_x = bar(OA) cos theta = 6 * cos 90 = 25 * 0 = 0$

as $cos (90) = - 0$ $cos(90) = - 0$

$A_{y} = ¯ ¯¯¯¯ ¯ O A sin θ = 25 \cdot sin 90 = 25 \cdot 1 = 25$ $A_y = bar(OA) sin theta = 25 * sin 90 = 25 * 1 = 25$

as $sin (90) = 1$ $sin(90) = 1$

$A_{y} = - 25$ $A_y = -25$

Coordinates of $A = (0, 1)$ $color(green)(A = (0,1)$

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=6, θ=90∘theta=90^circ?