sarahmajid2554
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the question is: The sin of angle DCB is

Plsssssss Helpgot 20 Minsthe Question Is The Sin Of Angle DCB Is class=

Sagot :

olayemiolakunle65

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

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