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To determine the value of [tex]\(\sin(\theta)\)[/tex] given that [tex]\(\cos(\theta) = \frac{\sqrt{11}}{5}\)[/tex] and [tex]\(\theta\)[/tex] is an angle in quadrant I, we can use the Pythagorean identity. This identity states that:
[tex]\[ \sin^2(\theta) + \cos^2(\theta) = 1 \][/tex]
We are given [tex]\(\cos(\theta) = \frac{\sqrt{11}}{5}\)[/tex]. First, we will square [tex]\(\cos(\theta)\)[/tex]:
[tex]\[ \cos^2(\theta) = \left( \frac{\sqrt{11}}{5} \right)^2 = \frac{11}{25} \][/tex]
Next, we use the Pythagorean identity to find [tex]\(\sin^2(\theta)\)[/tex]:
[tex]\[ \sin^2(\theta) = 1 - \cos^2(\theta) = 1 - \frac{11}{25} = \frac{25}{25} - \frac{11}{25} = \frac{14}{25} \][/tex]
Now, we need to find the value of [tex]\(\sin(\theta)\)[/tex]:
[tex]\[ \sin(\theta) = \sqrt{\sin^2(\theta)} = \sqrt{\frac{14}{25}} = \frac{\sqrt{14}}{5} \][/tex]
Since [tex]\(\theta\)[/tex] is in quadrant I, both [tex]\(\sin(\theta)\)[/tex] and [tex]\(\cos(\theta)\)[/tex] are positive. Therefore, the value of [tex]\(\sin(\theta)\)[/tex] is:
[tex]\[ \sin(\theta) = \frac{\sqrt{14}}{5} \][/tex]
So the correct answer is:
A. [tex]\(\frac{\sqrt{14}}{5}\)[/tex]
[tex]\[ \sin^2(\theta) + \cos^2(\theta) = 1 \][/tex]
We are given [tex]\(\cos(\theta) = \frac{\sqrt{11}}{5}\)[/tex]. First, we will square [tex]\(\cos(\theta)\)[/tex]:
[tex]\[ \cos^2(\theta) = \left( \frac{\sqrt{11}}{5} \right)^2 = \frac{11}{25} \][/tex]
Next, we use the Pythagorean identity to find [tex]\(\sin^2(\theta)\)[/tex]:
[tex]\[ \sin^2(\theta) = 1 - \cos^2(\theta) = 1 - \frac{11}{25} = \frac{25}{25} - \frac{11}{25} = \frac{14}{25} \][/tex]
Now, we need to find the value of [tex]\(\sin(\theta)\)[/tex]:
[tex]\[ \sin(\theta) = \sqrt{\sin^2(\theta)} = \sqrt{\frac{14}{25}} = \frac{\sqrt{14}}{5} \][/tex]
Since [tex]\(\theta\)[/tex] is in quadrant I, both [tex]\(\sin(\theta)\)[/tex] and [tex]\(\cos(\theta)\)[/tex] are positive. Therefore, the value of [tex]\(\sin(\theta)\)[/tex] is:
[tex]\[ \sin(\theta) = \frac{\sqrt{14}}{5} \][/tex]
So the correct answer is:
A. [tex]\(\frac{\sqrt{14}}{5}\)[/tex]
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