Georgina is told that an angle has a sine of (√(5))/(5) and is in Quadrant II. Use the Pythagorean identity \sin^2(\theta) + \cos^2(\theta) = 1, the trigonometric identity \tan \theta = (\sin \theta)/(\cos \theta), and the quadrant to find the tangent.

Question

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Answer 1

To find the tangent of the angle given that \(\sin \theta = \frac{\sqrt{5}}{5}\) and that the angle \(\theta\) is in Quadrant II, we will proceed with the following steps:

1. Use the Pythagorean identity: \[ \sin^2(\theta) + \cos^2(\theta) = 1 \]

   First, we calculate \(\sin^2(\theta)\): \[ \sin^2(\theta) = \left( \frac{\sqrt{5}}{5} \right)^2 = \frac{5}{25} = \frac{1}{5} \]

   Now, substituting into the Pythagorean identity: \[ \frac{1}{5} + \cos^2(\theta) = 1 \]

   Rearranging gives us: \[ \cos^2(\theta) = 1 - \frac{1}{5} = \frac{4}{5} \]

   To find \(\cos(\theta)\), we take the square root: \[ \cos(\theta) = \pm \sqrt{\frac{4}{5}} = \pm \frac{2}{\sqrt{5}} = \pm \frac{2\sqrt{5}}{5} \]

   Since \(\theta\) is in Quadrant II, where cosine is negative, we have: \[ \cos(\theta) = -\frac{2\sqrt{5}}{5} \]

2. Find the tangent of the angle: \[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \] Substituting the values of \(\sin(\theta)\) and \(\cos(\theta)\): \[ \tan(\theta) = \frac{\frac{\sqrt{5}}{5}}{-\frac{2\sqrt{5}}{5}} = \frac{\sqrt{5}}{5} \cdot \frac{-5}{2\sqrt{5}} = -\frac{1}{2} \]

Thus, the tangent of the angle is: \[ \boxed{-\frac{1}{2}} \]