Math Problem Statement

determine the angle in which angle theta is located if cos theta < 0 and tan theta > 0

Solution

To determine the quadrant in which angle θ\theta is located given that cosθ<0\cos \theta < 0 and tanθ>0\tan \theta > 0, let's analyze the properties of the trigonometric functions in each quadrant:

Properties of cosθ\cos \theta and tanθ\tan \theta in Different Quadrants:

  1. Quadrant I:

    • cosθ>0\cos \theta > 0
    • tanθ>0\tan \theta > 0
  2. Quadrant II:

    • cosθ<0\cos \theta < 0
    • tanθ<0\tan \theta < 0
  3. Quadrant III:

    • cosθ<0\cos \theta < 0
    • tanθ>0\tan \theta > 0
  4. Quadrant IV:

    • cosθ>0\cos \theta > 0
    • tanθ<0\tan \theta < 0

Analyzing the Conditions:

  • The condition cosθ<0\cos \theta < 0 tells us that the angle θ\theta must be in either Quadrant II or Quadrant III because cosine is negative in these two quadrants.
  • The condition tanθ>0\tan \theta > 0 tells us that the angle θ\theta must be in either Quadrant I or Quadrant III because tangent is positive in these two quadrants.

Conclusion:

The only quadrant that satisfies both conditions (cosθ<0\cos \theta < 0 and tanθ>0\tan \theta > 0) is Quadrant III.

So, the angle θ\theta is located in Quadrant III.

Would you like any more details or further clarifications?


Here are 5 related questions you might find useful:

  1. What are the signs of sine and cosine in each quadrant?
  2. How does the sign of the tangent function relate to the signs of sine and cosine?
  3. How would you determine the quadrant if sinθ>0\sin \theta > 0 and tanθ<0\tan \theta < 0?
  4. What happens to the trigonometric function values if θ\theta is shifted by 180 degrees?
  5. Can you determine the exact angle θ\theta if cosθ=12\cos \theta = -\frac{1}{2} and sinθ=32\sin \theta = -\frac{\sqrt{3}}{2}?

Tip: Remember that the tangent of an angle is positive where sine and cosine have the same sign, and negative where they have opposite signs.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Quadrants

Formulas

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Theorems

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Suitable Grade Level

High School