Question

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

Answer 1

i. <DCB =
`53.13^(o)`

ii. Sin of <DCB = 0.8

Explanation:

Let <DCB be represented by θ, so that;

Sin θ =
`(opposite)/(hypotenuse)`

Thus from the given diagram, we have;

Sin θ =
`(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