Answer:
sinC = [tex]\frac{4\sqrt{41} }{41}[/tex]
Step-by-step explanation:
We require to find the side DE before finding sinC
Using Pythagoras' identity in the right triangle
DE² + CE² = CD²
DE² + 5² = ([tex]\sqrt{41}[/tex] )²
DE² + 25 = 41 ( subtract 25 from both sides )
DE² = 16 ( take the square root of both sides )
DE = [tex]\sqrt{16}[/tex] = 4 , then
sinC = [tex]\frac{opposite}{hypotenuse}[/tex] = [tex]\frac{DE}{CD}[/tex] = [tex]\frac{4}{\sqrt{41} }[/tex] , rationalise the denominator
sinC = [tex]\frac{4}{\sqrt{41} }[/tex] × [tex]\frac{\sqrt{41} }{\sqrt{41} }[/tex] = [tex]\frac{4\sqrt{41} }{41}[/tex]
