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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?

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Trigonometric ratios

$$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$$

where:

- $$\theta$$ is the angle
- O is the side opposite the angle
- A is the side adjacent the angle
- H is the hypotenuse (the side opposite the right angle)

To find sin x and cos y, we must first find the length of the hypotenuse (PO) of the given right triangle.

To do this, use Pythagoras' Theorem:

$$a^2+b^2=c^2 \quad \textsf{(where a and b are the legs, and c is the hypotenuse)}$$

Given:

- a = 6
- b = 8
- c = PO

$$\implies 6^2+8^2=PO^2$$

$$\implies PO^2=100$$

$$\implies PO=\sqrt{100}$$

$$\implies PO=10$$

Using the found value of the hypotenuse and the trig ratios quoted above:

$$\implies \sf \sin(x)=\dfrac{6}{10}=\dfrac{3}{5}$$

$$\implies \sf \cos(y)=\dfrac{6}{10}=\dfrac{3}{5}$$

Therefore, $$\sf \sin(x)=\cos(y)$$
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Find Hypotenuse

- √6²+8²
- √10²
- 10

$$\\ \rm\Rrightarrow sinx=\dfrac{Perpendicular}{Hypotenuse}$$

$$\\ \rm\Rrightarrow sinx=\dfrac{6}{10}$$

$$\\ \rm\Rrightarrow sinx=\dfrac{3}{5}$$

And

$$\\ \rm\Rrightarrow cosy=\dfrac{Base}{Hypotenuse}$$

$$\\ \rm\Rrightarrow cosy=\dfrac{6}{10}$$

$$\\ \rm\Rrightarrow cosy=\dfrac{3}{5}$$
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