What is the answer to this
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[tex]S\frac{O}{H} C\frac{A}{H} T\frac{O}{A}[/tex] or [tex]Sin\frac{Opposite}{Hypotenuse}Cos\frac{Adjacent}{Hypotenuse} Tan\frac{Opposite}{Adjacent}[/tex]
At the point/angle E:
- the adjacent side(side next to the point/angle) is ED
- the opposite side(side of the triangle across from the point/angle) is DF
- the hypotenuse (the longest side of the triangle) is EF.
sin ∠E = [tex]\frac{opposite}{hypotenuse}[/tex]
sin ∠E = [tex]\frac{6}{10} = \frac{3}{5}[/tex] (simplified)
sinE = [tex]\frac{3}{5}[/tex]
the sine ratio is defined as
sin( angle ) = [tex]\frac{opposite}{hypotenuse}[/tex]
here the opposite to ∠E is 6 and the hypotenuse is 10
sinE = [tex]\frac{6}{10}[/tex] = [tex]\frac{3}{5}[/tex]