Question%0D%0ASteph is told that an angle has a sine of 14%0D%0A1%0D%0A4%0D%0A and is in Quadrant II. Use the Pythagorean identity sin2θ+cos2θ=1%0D%0Asin%0D%0A2%0D%0A%E2%81%A1%0D%0A𝜃%0D%0A+%0D%0Acos%0D%0A2%0D%0A%E2%81%A1%0D%0A𝜃%0D%0A=%0D%0A1%0D%0A, the trigonometric identity tanθ=sinθcosθ%0D%0Atan%0D%0A%E2%81%A1%0D%0A𝜃%0D%0A=%0D%0Asin%0D%0A%E2%81%A1%0D%0A𝜃%0D%0Acos%0D%0A%E2%81%A1%0D%0A𝜃%0D%0A, and the quadrant to find the tangent. (1 point)

Answer 1

To find the tangent of the angle \( \theta \) given that \( \sin \theta = 0.14 \) and that \( \theta \) is in Quadrant II, we can use the Pythagorean identity and the definitions of trigonometric functions.

1. Use the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting \( \sin \theta = 0.14 \): \[ (0.14)^2 + \cos^2 \theta = 1 \] \[ 0.0196 + \cos^2 \theta = 1 \] \[ \cos^2 \theta = 1 - 0.0196 = 0.9804 \] \[ \cos \theta = \pm \sqrt{0.9804} \approx \pm 0.99 \]

   Since \( \theta \) is in Quadrant II, where cosine is negative, we have: \[ \cos \theta \approx -0.99 \]

2. Find the tangent using the identity: The tangent of the angle \( \theta \) is given by: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] Substituting the values we found: \[ \tan \theta = \frac{0.14}{-0.99} \approx -0.1414 \]

So, the tangent of the angle \( \theta \) is: \[ \tan \theta \approx -0.1414 \]