Find the value of x and y in the right angled triangle;
[tex]\begin{gathered} \text{Opposite}=y\text{ (facing the reference angle)} \\ \text{Adjacent}=x\text{ (lies between the reference angle and the right angle)} \\ \text{Hypotenuse}=16\sqrt[]{3}\text{ (facing the right angle)} \end{gathered}[/tex]
To calculate x, we shall use the trig ratio;
[tex]\begin{gathered} \cos \theta=\frac{adj}{hyp} \\ \cos 30=\frac{x}{16\sqrt[]{3}} \\ \text{Note that cos 30 is also equal to }\frac{\sqrt[]{3}}{2}.\text{ Therefore} \\ \frac{\sqrt[]{3}}{2}=\frac{x}{16\sqrt[]{3}} \\ \text{Cross multiply and you'll have} \\ \frac{16\sqrt[]{3}\times\sqrt[]{3}}{2}=x \\ \frac{16\times3}{2}=x \\ x=24 \end{gathered}[/tex]
To valculate y, we shall use the trig ratio;
[tex]\begin{gathered} \sin \theta=\frac{opp}{hyp} \\ \sin 30=\frac{y}{16\sqrt[]{3}} \\ Note\text{ that sin 30 is also equal to }\frac{1}{2}.\text{ Therefore} \\ \frac{1}{2}=\frac{y}{16\sqrt[]{3}} \\ \text{Cross multiply and you'll have} \\ \frac{16\sqrt[]{3}}{2}=y \\ y=8\sqrt[]{3} \end{gathered}[/tex]
The first option is the correct answer
x = 24, y = 8 square root 3
