To solve for \(\cos y^\circ\) in the context of the given equations involving \(\sin y^\circ\) and \(\tan y^\circ\):
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From the given \(\sin y^\circ = \frac{a}{6}\), we can derive the opposite side in the triangle as \(a\) and the hypotenuse as \(6\).
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For \(\tan y^\circ\), we have \(\tan y^\circ = \frac{a}{b}\). Recall that \(\tan y^\circ\) is also defined as \(\frac{\text{opposite}}{\text{adjacent}}\). This indicates that the adjacent side is \(b\).
Now, we apply the Pythagorean theorem to find the adjacent side \(b\):
\[ \text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2 \] \[ 6^2 = a^2 + b^2 \] \[ 36 = a^2 + b^2 \]
Next, we need to find \(\cos y^\circ\). The definition of cosine in a right triangle, given an angle \(y^\circ\), is:
\[ \cos y^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{b}{6} \]
Among the options provided, the correct expression for \(\cos y^\circ\) is:
\[ \cos y^\circ = \frac{b}{6} \]
Therefore, the value of \(\cos y^\circ\) is:
\[ \boxed{\frac{b}{6}} \]
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