A robotic arm starts at the point (1,0) on the unit circle and rotates counterclockwise by… — ACT

A robotic arm starts at the point $(1,0)$ on the unit circle and rotates counterclockwise by $\frac{11\pi}{6}$ radians. It then rotates clockwise by $120^\circ$ . At which of the following coordinates on the unit circle does the arm's tip end?

$\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$ Answer
B
$\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$
C
$(0, 1)$
D
$(-1, 0)$
E
$\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$

Answer

\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)

The correct answer is obtained by converting the initial rotation of

\frac{11\pi}{6}

radians to

330^\circ

, subtracting the clockwise rotation of

120^\circ

to get

210^\circ

, and finding the cosine and sine values for this angle in the third quadrant, which are

-\frac{\sqrt{3}}{2}

and

-\frac{1}{2}

respectively.

Step-by-Step Solution

Convert the initial counterclockwise rotation from radians to degrees.

\frac{11\pi}{6} \text{ radians} \times \frac{180^\circ}{\pi} = 11 \times 30^\circ = 330^\circ

Converting both angles to degrees makes them easier to combine.

Apply the second rotation (clockwise, which means subtracting the angle).

330^\circ - 120^\circ = 210^\circ

Clockwise rotation reduces the angle in standard position.

Find the unit circle coordinates for the final angle of

210^\circ

x = \cos(210^\circ) = -\frac{\sqrt{3}}{2}

and

y = \sin(210^\circ) = -\frac{1}{2}

, yielding the coordinates

\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)

The terminal ray of

210^\circ

lies in Quadrant III, where both sine and cosine are negative, with a reference angle of

30^\circ

Key Concept

To find coordinates of a rotated point on the unit circle, convert the angle measures to a common unit, compute the net rotation angle in standard position, and evaluate the cosine (for the

x

-coordinate) and sine (for the

y

-coordinate) of the resulting angle.

Estimated Time:1m 30s

Question

Answer

Step-by-Step Solution

Key Concept

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