To solve the problem, let's recall the definitions of the trigonometric functions involved.
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Given:
- \(\sin y^\circ = \frac{a}{6}\)
- \(\tan y^\circ = \frac{a}{b}\)
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Definitions of trigonometric functions:
- \(\sin y^\circ = \frac{\text{opposite}}{\text{hypotenuse}}\)
- \(\tan y^\circ = \frac{\text{opposite}}{\text{adjacent}}\)
- \(\sec y^\circ = \frac{1}{\cos y^\circ} = \frac{\text{hypotenuse}}{\text{adjacent}}\)
Now, from the definition, we know that the triangle has the following sides:
- Opposite side: \(a\)
- Hypotenuse: \(6\)
Using the Pythagorean theorem, we can find the adjacent side (\(c\)): \[ c^2 + a^2 = 6^2 \rightarrow c^2 + a^2 = 36 \rightarrow c^2 = 36 - a^2 \rightarrow c = \sqrt{36 - a^2} \]
Now, using the given information:
- \(\tan y^\circ = \frac{a}{b}\)
- Using the definition of tangent: \[ \tan y^\circ = \frac{a}{c} = \frac{a}{\sqrt{36 - a^2}} \rightarrow \frac{a}{b} = \frac{a}{\sqrt{36 - a^2}} \] This implies: \[ b = \sqrt{36 - a^2} \]
Now, we can find \(\sec y^\circ\): \[ \sec y^\circ = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{6}{c} \]
Substituting \(c\): \[ \sec y^\circ = \frac{6}{\sqrt{36 - a^2}} \]
Now substituting \(b = \sqrt{36 - a^2}\): \[ \sec y^\circ = \frac{6}{b} \]
Thus, the value of \(\sec y^\circ\) is expressed in terms of \(b\) as: \[ \sec y^\circ = \frac{6}{b} \]
Therefore, among the options provided, the correct answer is: \[ \sec y^\circ = \frac{6}{b} \]
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