Question

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...

• A. \(6\sqrt{3}+\frac{5}{2}\)
• B. \(\frac{5\sqrt{3}}{2}-6\)
• C. \(\frac{5\sqrt{3}}{2}+6\)
• D. \(6\sqrt{3}-\frac{5}{2}\)

Hint:

Rotating (x, y) by an angle t gives y-coordinate x sin t + y cos t.

Answer 1

The Correct Option is A

Step 1: Use complex numbers
Write A as $5 + 12i$. A rotation by $30^\circ$ anticlockwise multiplies by $e^{i\pi/6} = \frac{\sqrt{3}}{2} + \frac{i}{2}$.

Step 2: Multiply
$(5 + 12i)\left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right) = \frac{5\sqrt{3}}{2} + \frac{5i}{2} + 6\sqrt{3}\,i - 6$.

Step 3: Read off the imaginary part
Real part: $\frac{5\sqrt{3}}{2} - 6$. Imaginary part: $\frac{5}{2} + 6\sqrt{3}$. The ordinate is the imaginary part.

Step 4: Check
Rotation keeps the distance from the origin: $|5 + 12i| = 13$ and $|e^{i\pi/6}| = 1$, so the new point is also 13 from O. The abscissa is $\frac{5\sqrt{3}}{2} - 6$, which is the value in option (B), so (B) is the abscissa and not the ordinate.

Final Answer:
The ordinate of B is 6 sqrt(3) + 5/2. This is option (A). \[ \boxed{\text{(A) }6\sqrt{3}+\frac{5}{2}} \]