Find the missing lengths of the sides

Answer: option c
Step-by-step explanation:
You can use these identities:
[tex]sin\alpha=\frac{opposite}{hypotenuse}\\\\tan\alpha=\frac{opposite}{adjacent}[/tex]
Then, using the angle that measures 30 degrees, you know that:
[tex]\alpha=30\°\\opposite=8\\adjacent=b[/tex]
Substituting:
[tex]tan(30\°)=\frac{8}{b}[/tex]
Now you must solve for b:
[tex]b=\frac{8}{tan(30\°)}\\\\b=8\sqrt{3}[/tex]
Using the angle that measures 30 degrees, you know that:
[tex]\alpha=30\°\\opposite=8\\hypotenuse=c[/tex]
Substituting:
[tex]sin(30\°)=\frac{8}{c}[/tex]
Now you must solve for c:
[tex]c=\frac{8}{sin(30\°)}\\\\c=16[/tex]
ANSWER
The correct answer is C
EXPLANATION
The side adjacent to the 60° angle is 8 units.
The hypotenuse is c.
Using the cosine ratio, we have
[tex] \cos(60 \degree) = \frac{adjacent}{hypotenuse} [/tex]
[tex]\cos(60 \degree) = \frac{8}{c} [/tex]
[tex] \frac{1}{2}= \frac{8}{c} [/tex]
Cross multiply
[tex]c = 8 \times 2 = 16[/tex]
Also
[tex]\cos(30 \degree) = \frac{b}{c} [/tex]
[tex]\cos(30 \degree) = \frac{b}{16} [/tex]
[tex] \frac{ \sqrt{3} }{2} = \frac{b}{16} [/tex]
Multiply both sides by 16
[tex]b = 16 \times \frac{ \sqrt{3} }{2} [/tex]
[tex]b = 8 \sqrt{3} [/tex]
The correct answer is C