|
|
|
|
|
|
|
|
|
|
|
|
Putar $$$\left(3 \sqrt{2}, - \frac{\sqrt{2}}{4}\right)$$$ sebesar $$$45^{\circ}$$$ berlawanan arah jarum jam dengan pusat $$$\left(0, 0\right)$$$
Masukan Anda
Putar $$$\left(3 \sqrt{2}, - \frac{\sqrt{2}}{4}\right)$$$ sebesar sudut $$$45^{\circ}$$$ berlawanan arah jarum jam dengan pusat $$$\left(0, 0\right)$$$.
Solusi
Rotasi titik $$$\left(x, y\right)$$$ terhadap titik asal dengan sudut $$$\theta$$$ berlawanan arah jarum jam akan menghasilkan titik baru $$$\left(x \cos{\left(\theta \right)} - y \sin{\left(\theta \right)}, x \sin{\left(\theta \right)} + y \cos{\left(\theta \right)}\right)$$$.
Dalam kasus kita, $$$x = 3 \sqrt{2}$$$, $$$y = - \frac{\sqrt{2}}{4}$$$, dan $$$\theta = 45^{\circ}$$$.
Oleh karena itu, titik baru adalah $$$\left(3 \sqrt{2} \cos{\left(45^{\circ} \right)} - - \frac{\sqrt{2}}{4} \sin{\left(45^{\circ} \right)}, 3 \sqrt{2} \sin{\left(45^{\circ} \right)} + - \frac{\sqrt{2}}{4} \cos{\left(45^{\circ} \right)}\right) = \left(\frac{13}{4}, \frac{11}{4}\right).$$$
Jawaban
Titik baru adalah $$$\left(\frac{13}{4}, \frac{11}{4}\right) = \left(3.25, 2.75\right)$$$A.