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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

[tex]\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}[/tex]

where:

- [tex]\theta[/tex] 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:

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

Given:

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

[tex]\implies 6^2+8^2=PO^2[/tex]

[tex]\implies PO^2=100[/tex]

[tex]\implies PO=\sqrt{100}[/tex]

[tex]\implies PO=10[/tex]

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

[tex]\implies \sf \sin(x)=\dfrac{6}{10}=\dfrac{3}{5}[/tex]

[tex]\implies \sf \cos(y)=\dfrac{6}{10}=\dfrac{3}{5}[/tex]

Therefore, [tex]\sf \sin(x)=\cos(y)[/tex]
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Find Hypotenuse

- √6²+8²
- √10²
- 10

[tex]\\ \rm\Rrightarrow sinx=\dfrac{Perpendicular}{Hypotenuse}[/tex]

[tex]\\ \rm\Rrightarrow sinx=\dfrac{6}{10}[/tex]

[tex]\\ \rm\Rrightarrow sinx=\dfrac{3}{5}[/tex]

And

[tex]\\ \rm\Rrightarrow cosy=\dfrac{Base}{Hypotenuse}[/tex]

[tex]\\ \rm\Rrightarrow cosy=\dfrac{6}{10}[/tex]

[tex]\\ \rm\Rrightarrow cosy=\dfrac{3}{5}[/tex]
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