Answer:
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:
[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]
