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the question is: The sin of angle DCB is

Answer:
i. <DCB = [tex]53.13^{o}[/tex]
ii. Sin of <DCB = 0.8
Step-by-step explanation:
Let <DCB be represented by θ, so that;
Sin θ = [tex]\frac{opposite}{hypotenuse}[/tex]
Thus from the given diagram, we have;
Sin θ = [tex]\frac{4}{5}[/tex]
= 0.8
This implies that,
θ = [tex]Sin^{-1}[/tex] 0.8
= 53.1301
θ = [tex]53.13^{o}[/tex]
Therefore, <DCB = [tex]53.13^{o}[/tex].
So that,
Sin of <DCB = Sin [tex]53.13^{o}[/tex]
= 0.8
Sin of <DCB = 0.8