NEW POSITION AFTER ROTATIONAL TRANSFORMATION BY GIVEN DEGREE

Problem 1 :

Point A (9, 12) rotates around the origin O in the plane through 60 degree in the anticlockwise direction to the new position B. Find the coordinates of the point B.

Solution :

Finding radius from the initial position lies on the circle (9, 12).

r = √9² + 12²

r = √81+144

r = √225

r = 15

With radius 15 units and angle measure of 60 degree, the new position will be at

15 cos (θ + 60) = 15 [ cos θ cos 60 - sin θ sin 60]

= 15[cos θ (1/2) - sin θ(√3/2)]

= 15cos θ (1/2) - 15sin θ(√3/2)

= 9(1/2) - 12(√3/2)

= (3/2)(3 - 4√3)

15 sin (θ + 60) = 15 [ sin θ cos 60 + cos θ sin 60]

= 15[sin θ(1/2) + cos θ(√3/2)]

= 15 sin θ (1/2) + 15 cos θ(√3/2)

= 12(1/2) + 9(√3/2)

= 3(4 + 3√3)

So, the new position will be at [(3/2)(3 - 4√3), (3/2)(4 + 3√3)]

Problem 2 :

Determine the image of the point A(4, -3) after the rotation of 60 degree about the origin.

Solution :

Finding radius from the initial position lies on the circle (4, -3).

r = √4² + (-3)²

r = √16+9

r = √25

r = 5

5 cos (θ + 60) = 5 [ cos θ cos 60 - sin θ sin 60]

= 5[cos θ (1/2) - sin θ(√3/2)]

= 5cos θ (1/2) - 5sin θ(√3/2)

= 4(1/2) - (-3)(√3/2)

= (2, 3√3/2)

5 sin (θ + 60) = 5 [ sin θ cos 60 + cos θ sin 60]

= 5[sin θ(1/2) + cos θ(√3/2)]

= 5 sin θ (1/2) + 5 cos θ(√3/2)

= 4(1/2) + (-3)(√3/2)

= (2, -3√3/2)

So, the new position will be at [(3/2)(3 - 4√3), (3/2)(4 + 3√3)]