what is the location of the green point? hint: notice the orange angle is labeled using radians (?,?)

what is the location of the green point? hint: notice the orange angle is labeled using radians (?,?)

what is the location of the green point? hint: notice the orange angle is labeled using radians (?,?)

Answer

Explanation:

Step1: Simplify the angle

Subtract (2\pi) (a full - circle in radians) from (\frac{53\pi}{45}) until the angle is in the range ([0, 2\pi]). [ \begin{align*} \frac{53\pi}{45}- 2\pi&=\frac{53\pi}{45}-\frac{90\pi}{45}\ &=-\frac{37\pi}{45} \end{align*} ] Since (-\frac{37\pi}{45}+2\pi=\frac{- 37\pi + 90\pi}{45}=\frac{53\pi}{45}), we can also use the formula for the coordinates of a point on a circle (x = r\cos\theta), (y = r\sin\theta). Here (r = 4) (the radius of the circle) and (\theta=\frac{53\pi}{45}). Another way: (\frac{53\pi}{45}= \pi+\frac{8\pi}{45}) (\cos(\frac{53\pi}{45})=\cos(\pi+\frac{8\pi}{45})=-\cos(\frac{8\pi}{45})), (\sin(\frac{53\pi}{45})=\sin(\pi+\frac{8\pi}{45})=-\sin(\frac{8\pi}{45})) Using a calculator: (\cos(\frac{8\pi}{45})\approx0.891), (\sin(\frac{8\pi}{45})\approx0.453) [ \begin{align*} x&=4\cos(\frac{53\pi}{45})\ &=4\times(- 0.891)\ &\approx - 3.564 \end{align*} ] [ \begin{align*} y&=4\sin(\frac{53\pi}{45})\ &=4\times(-0.453)\ &\approx - 1.812 \end{align*} ] Since we are likely to assume a more "exact" - like (rounded to two - decimal - places for grid - like) calculation (and using the unit - circle concept extended to radius (r = 4)): We know that for a point ((x,y)) on a circle of radius (r) with angle (\theta) (in radians) from the positive (x) - axis, (x = r\cos\theta) and (y=r\sin\theta). [ \begin{align*} x&=4\cos(\frac{53\pi}{45})\ &=4\cos( \pi+\frac{8\pi}{45})\ &=-4\cos(\frac{8\pi}{45})\approx - 4\times0.89\approx - 3.56 \end{align*} ] [ \begin{align*} y&=4\sin(\frac{53\pi}{45})\ &=4\sin(\pi+\frac{8\pi}{45})\ &=-4\sin(\frac{8\pi}{45})\approx - 4\times0.45\approx - 1.8 \end{align*} ]

Answer:

((-4\cos(\frac{8\pi}{45}),-4\sin(\frac{8\pi}{45}))\approx(-3.56, - 1.8))