Question

Plsssssss helpgot 20 mins​the question is: The sin of angle DCB is

Answer 1

Answer:

i. <DCB = $53.13^{o}$

ii. Sin of <DCB = 0.8

Step-by-step explanation:

Let <DCB be represented by θ, so that;

Sin θ = $\frac{opposite}{hypotenuse}$

Thus from the given diagram, we have;

Sin θ = $\frac{4}{5}$

        = 0.8

This implies that,

θ = $Sin^{-1}$ 0.8

   = 53.1301

θ = $53.13^{o}$

Therefore, <DCB = $53.13^{o}$.

So that,

Sin of <DCB = Sin $53.13^{o}$

                    = 0.8

Sin of <DCB = 0.8