Q51 / 150 · MHT CET 2026 (19 April Shift 2) · Mathematics
Point A$(5, 12)$ rotated about the origin O in the XY-plane through an angle of $30^\circ$ in the anticlockwise direction to a new position B. The ordinate of point B is...
- Correct answerA$6\sqrt{3} + \dfrac{5}{2}$
- B$\dfrac{5\sqrt{3}}{2} - 6$
- C$\dfrac{5\sqrt{3}}{2} + 6$
- D$6\sqrt{3} - \dfrac{5}{2}$
Solution
Let the coordinates of point A be $(x, y) = (5, 12)$.
When a point $(x, y)$ is rotated about the origin by an angle $\theta$ in the anticlockwise direction, the new coordinates $(x', y')$ are given by:
$x' = x \cos\theta - y \sin\theta$
$y' = x \sin\theta + y \cos\theta$
Substituting $x = 5$, $y = 12$, and $\theta = 30^\circ$ to find the ordinate $y'$ of the new point B:
$y' = 5 \sin 30^\circ + 12 \cos 30^\circ$
$y' = 5 \left(\dfrac{1}{2}\right) + 12 \left(\dfrac{\sqrt{3}}{2}\right)$
$y' = \dfrac{5}{2} + 6\sqrt{3}$
The ordinate of point B is $6\sqrt{3} + \dfrac{5}{2}$.
Answer: $6\sqrt{3} + \dfrac{5}{2}$
When a point $(x, y)$ is rotated about the origin by an angle $\theta$ in the anticlockwise direction, the new coordinates $(x', y')$ are given by:
$x' = x \cos\theta - y \sin\theta$
$y' = x \sin\theta + y \cos\theta$
Substituting $x = 5$, $y = 12$, and $\theta = 30^\circ$ to find the ordinate $y'$ of the new point B:
$y' = 5 \sin 30^\circ + 12 \cos 30^\circ$
$y' = 5 \left(\dfrac{1}{2}\right) + 12 \left(\dfrac{\sqrt{3}}{2}\right)$
$y' = \dfrac{5}{2} + 6\sqrt{3}$
The ordinate of point B is $6\sqrt{3} + \dfrac{5}{2}$.
Answer: $6\sqrt{3} + \dfrac{5}{2}$
Hence, the correct answer is option: 'A'.