Math Problem Statement

Find the reference angle for the given angle. (a)
9𝜋 10

Correct: Your answer is correct.

(b)
11𝜋 8

Incorrect: Your answer is incorrect.

(c)
10𝜋 3

Solution

To find the reference angle, we need to understand the position of the given angle in relation to the nearest x-axis and then find the acute angle that it makes with the x-axis. Here's how to handle each case:

(a) 9π10\frac{9\pi}{10}

  1. This is already in radians. Since it's less than π\pi (which is 10π10\frac{10\pi}{10}), it's located in the second quadrant.
  2. To find the reference angle for angles in the second quadrant, we subtract the given angle from π\pi. Reference Angle=π9π10=10π109π10=π10\text{Reference Angle} = \pi - \frac{9\pi}{10} = \frac{10\pi}{10} - \frac{9\pi}{10} = \frac{\pi}{10} So, the reference angle is π10\frac{\pi}{10}.

(b) 11π8\frac{11\pi}{8}

  1. Since 11π8>π\frac{11\pi}{8} > \pi, this angle is in the third quadrant. For third quadrant angles, the reference angle is found by subtracting π\pi from the given angle. Reference Angle=11π8π=11π88π8=3π8\text{Reference Angle} = \frac{11\pi}{8} - \pi = \frac{11\pi}{8} - \frac{8\pi}{8} = \frac{3\pi}{8} Thus, the reference angle is 3π8\frac{3\pi}{8}.

(c) 10π3\frac{10\pi}{3}

  1. Since 10π3\frac{10\pi}{3} is greater than 2π2\pi, we subtract 2π2\pi (i.e., 6π3\frac{6\pi}{3}) to bring it within one full revolution. 10π32π=10π36π3=4π3\frac{10\pi}{3} - 2\pi = \frac{10\pi}{3} - \frac{6\pi}{3} = \frac{4\pi}{3}
  2. 4π3\frac{4\pi}{3} is in the third quadrant. For third quadrant angles, we subtract π\pi from the angle: Reference Angle=4π3π=4π33π3=π3\text{Reference Angle} = \frac{4\pi}{3} - \pi = \frac{4\pi}{3} - \frac{3\pi}{3} = \frac{\pi}{3} Thus, the reference angle is π3\frac{\pi}{3}.

Summary:

  • (a) 9π10\frac{9\pi}{10}: Reference angle is π10\frac{\pi}{10}.
  • (b) 11π8\frac{11\pi}{8}: Reference angle is 3π8\frac{3\pi}{8}.
  • (c) 10π3\frac{10\pi}{3}: Reference angle is π3\frac{\pi}{3}.

Would you like further clarification on any of these? Here are five questions that extend the concept of reference angles:

  1. How would you find the reference angle for 13π4\frac{13\pi}{4}?
  2. What is the reference angle for 7π6-\frac{7\pi}{6}?
  3. Can you explain why reference angles are always positive and acute?
  4. How do reference angles relate to the unit circle?
  5. How does the reference angle change for angles in the fourth quadrant?

Tip: When calculating reference angles, always reduce angles greater than 2π2\pi by subtracting 2π2\pi to find the equivalent angle in the first full rotation.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Reference Angles
Radians

Formulas

Reference angle formula for angles in second quadrant: Reference Angle = π - θ
Reference angle formula for angles in third quadrant: Reference Angle = θ - π
Reference angle formula for angles greater than 2π: Reference Angle = θ - 2π

Theorems

Reference Angle Theorem

Suitable Grade Level

Grades 10-12