Answer:
The answet is C.
Step-by-step explanation:
First, you have to find the angle of ACB using Sine Rule, sinθ = opposite/hypotenuse :
[tex] \sin(θ ) = \frac{oppo.}{hypo.} [/tex]
[tex]let \: oppo. = 4 \\ let \: hypo. = 5[/tex]
[tex] \sin(θ) = \frac{4}{5} [/tex]
[tex]θ = {\sin( \frac{4}{5} ) }^{ - 1} [/tex]
[tex]θ = 53.1 \: (1d.p)[/tex]
Given that line AB is parallel to line CD so ∠C = 90°. Next, you have to find the angle of ACD :
[tex]ACD = 90 - 53.1 = 36.9[/tex]
Lastly, you can find the length of CD using Cosine rule, cosθ = adjacent/hypotenuse :
[tex] \cos(θ) = \frac{adj.}{hypo.} [/tex]
[tex]let \: θ = 36.9 \\ let \: adj. = 5 \\ let \: hypo. = CD[/tex]
[tex] \cos(36.9 ) = \frac{5}{CD} [/tex]
[tex]CD \cos(36.9) = 5[/tex]
[tex]CD = \frac{5}{ \cos(36.9) } [/tex]
[tex]CD = 6.25 units\: (3s.f)[/tex]
