Where will the terminal side of \( \frac{41\pi}{6} \) lie?
Answer and explanation
Correct answer: Second quadrant
Subtracting three complete revolutions, where \(2\pi=\frac{12\pi}{6}\), gives \(\frac{41\pi}{6}-\frac{36\pi}{6}=\frac{5\pi}{6}\). Now \(\frac{5\pi}{6}=150^\circ\), which lies between \(\frac{\pi}{2}\) and \(\pi\); hence its terminal side lies in the second quadrant. The fourth quadrant contains angles between \(\frac{3\pi}{2}\) and \(2\pi\). Exam tip: reduce a large angle by subtracting multiples of \(2\pi\) before identifying its quadrant.
Frequently asked questions
What is the correct answer to this question?
Second quadrant
Why is this the correct answer?
Subtracting three complete revolutions, where \(2\pi=\frac{12\pi}{6}\), gives \(\frac{41\pi}{6}-\frac{36\pi}{6}=\frac{5\pi}{6}\). Now \(\frac{5\pi}{6}=150^\circ\), which lies between \(\frac{\pi}{2}\) and \(\pi\); hence its terminal side lies in the second quadrant. The fourth quadrant contains angles between \(\frac{3\pi}{2}\) and \(2\pi\). Exam tip: reduce a large angle by subtracting multiples of \(2\pi\) before identifying its quadrant.
Which subject and chapter does this question cover?
This is a Class 12 Mathematics question. Chapter: Trigonometric Functions. Topic: Angles.
