Question

Q27. Perform 60 degree rotation of a triangle with vertices (0, 0) (1, 1) (5, 2) about origin. ( मूलबिंदु के बारे में (0, 0) (1, 1) (5, 2) के साथ त्रिकोण के 60 डिग्री रोटेशन का प्रदर्शन करें।)

Answer 1

Answer:

please write the question in English with proper dictionary

Answer 2

Answer:

The new vertices are (0, 0), ($\frac{1-\sqrt{3} }{2}, \frac{1+\sqrt{3} }{2}$), and ($\frac{5-2\sqrt{3} }{2}, \frac{2+5\sqrt{3} }{2}$).

Step-by-step explanation:

Step 1:

Let us represent the given coordinates in matrix form using homogeneous coordinates of the vertices.

[A B C] = $\left[\begin{array}{cccc}A&0&0&1\\B&1&1&1\\C&5&2&1\end{array}\right]$

Step 2:

Let the matrix of rotation be R₆₀ as a 60° rotation is to be performed.

R₆₀ = $\left[\begin{array}{ccc}cos(60)&sin(60)&0\\-sin(60)&cos(60)&0\\0&0&1\end{array}\right]$

R₆₀ = $\left[\begin{array}{ccc}\frac{1}{2} &\frac{\sqrt{3} }{2} &0\\-\frac{\sqrt{3} }{2} &\frac{1}{2} &0\\0&0&1\end{array}\right]$

Step 3:

The new coordinates A'B'C' of the rotated triangle ABC can be given as

[A'B'C'] = [ABC]R₆₀

$[A'B'C'] = \left[\begin{array}{ccc}0&0&1\\1&1&1\\5&2&1\end{array}\right] \left[\begin{array}{ccc}\frac{1}{2} &\frac{\sqrt{3} }{2} &0\\-\frac{\sqrt{3} }{2} &\frac{1}{2} &0\\0&0&1\end{array}\right]$

Using matrix multiplication, we get

$[A'B'C'] = \left[\begin{array}{ccc}0&0&1\\\frac{1-\sqrt{3}}{2} &\frac{1+\sqrt{3}}{2} &1\\\frac{5-2\sqrt{3}}{2} &\frac{2+5\sqrt{3}}{2} &1\end{array}\right]$

Therefore, the new coordinates of the triangle are (0, 0), ($\frac{1-\sqrt{3} }{2}, \frac{1+\sqrt{3} }{2}$), and ($\frac{5-2\sqrt{3} }{2}, \frac{2+5\sqrt{3} }{2}$).

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