Question

question 2 (essay worth 10 points)
(05 02 mc)
use the image below to answer the following question. find the value of sin x° and cos y°. what relationship do the ratios of sin x° and cos y° share?

Explanation:

Step1: Find the hypotenuse

Use the Pythagorean theorem \(c=\sqrt{a^{2}+b^{2}}\), where \(a = 6\) and \(b=8\).
\(c=\sqrt{6^{2}+8^{2}}=\sqrt{36 + 64}=\sqrt{100}=10\)

Step2: Calculate \(\sin x^{\circ}\)

By the definition of sine in a right - triangle \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). For angle \(x\), the opposite side is \(6\) and the hypotenuse is \(10\).
\(\sin x^{\circ}=\frac{6}{10}=\frac{3}{5}\)

Step3: Calculate \(\cos y^{\circ}\)

By the definition of cosine in a right - triangle \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). For angle \(y\), the adjacent side is \(6\) and the hypotenuse is \(10\).
\(\cos y^{\circ}=\frac{6}{10}=\frac{3}{5}\)

Step4: Analyze the relationship

Since \(\sin x^{\circ}=\frac{3}{5}\) and \(\cos y^{\circ}=\frac{3}{5}\), we can conclude that \(\sin x^{\circ}=\cos y^{\circ}\)

Answer:

\(\sin x^{\circ}=\frac{3}{5}\), \(\cos y^{\circ}=\frac{3}{5}\), and \(\sin x^{\circ}=\cos y^{\circ}\)