“angle A = 32, AC = 9, BC = x. Find x to two decimal places, don’t forget to round properly based on the third decimal place.” Pls answer & explain thanks

Given:
Given that ABC is a right triangle.
The measure of ∠A is 32°.
The length of AC is 9 units.
The length of BC is x units.
We need to determine the value of x.
Value of x:
The value of x can be determined using the trigonometric ratio.
Thus, we have;
[tex]sin \ \theta=\frac{opp}{hyp}[/tex]
where θ = A, the side opposite to A is BC and hypotenuse is AC.
Thus, we have;
[tex]sin \ A=\frac{BC}{AC}[/tex]
Substituting BC = x and AC = 9, we get;
[tex]sin \ 32=\frac{x}{9}[/tex]
Multiplying both sides by 9, we have;
[tex]sin \ 32 \times 9=x[/tex]
[tex]4.77=x[/tex]
Thus, the value of x is 4.77 units.