Question

rotate the point 95° to within 10°.
(0,1)
(-1,0) (1,0)
(0,-1)
answer attempt 1 out of 3
you must answer all questions above in order to submit.

Explanation:

Step1: Identify the initial point

The initial point is \((1, 0)\) on the unit circle (since it's at the rightmost point of the circle with radius 1, centered at the origin).

Step2: Recall rotation of points on unit circle

For a point \((x, y)\) on the unit circle, rotating by an angle \(\theta\) counterclockwise gives the point \((\cos\theta, \sin\theta)\). Here, we start at \((1, 0)\) which corresponds to an angle of \(0^\circ\), and we rotate \(95^\circ\) counterclockwise.

Step3: Calculate the coordinates after rotation

Using the unit circle rotation formula, the new coordinates \((x, y)\) after rotating \(95^\circ\) from \((1, 0)\) (which is \(0^\circ\)) are \((\cos(95^\circ), \sin(95^\circ))\). Let's compute these values:

• \(\cos(95^\circ) \approx \cos(90^\circ + 5^\circ)=-\sin(5^\circ)\approx - 0.0872\)
• \(\sin(95^\circ)=\sin(90^\circ + 5^\circ)=\cos(5^\circ)\approx0.9962\)

So the point after rotating \(95^\circ\) (within \(10^\circ\) accuracy, so \(95^\circ\) itself is within the tolerance) from \((1, 0)\) is approximately \((\cos(95^\circ), \sin(95^\circ))\approx(-0.087, 0.996)\). If we consider the angle, the angle of the new point from the positive x - axis is \(95^\circ\), so the point lies in the second quadrant, close to the \(90^\circ\) (or \(\frac{\pi}{2}\) radians) position but \(5^\circ\) into the second quadrant.

Answer:

The coordinates of the point after rotating \(95^\circ\) (within \(10^\circ\) of \(95^\circ\)) from \((1, 0)\) are approximately \(\boldsymbol{(-0.09, 0.996)}\) (or in terms of the angle - defined point on the unit circle at \(95^\circ\)). If we consider the angle - based description, the point is at an angle of \(95^\circ\) from the positive x - axis on the unit circle, with coordinates \((\cos(95^\circ), \sin(95^\circ))\).