Question

find the length of side ( x ) to the nearest tenth.

Explanation:

Step1: Identify triangle type and trigonometric ratio

This is a right - triangle. We can use the sine function. The sine of an angle in a right - triangle is defined as $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$. For the $60^{\circ}$ angle, the side opposite to it is 12, and the hypotenuse is $x$. Also, we know that $\sin(60^{\circ})=\frac{\sqrt{3}}{2}\approx0.866$.
We can also use the sine of the $30^{\circ}$ angle. The side opposite to $30^{\circ}$ is the shorter leg. But let's use the angle - side relationship. Let's consider the angle of $60^{\circ}$. The side opposite to $60^{\circ}$ is 12, and the hypotenuse is $x$. So $\sin(60^{\circ})=\frac{12}{x}$.

Step2: Solve for x

From $\sin(60^{\circ})=\frac{12}{x}$, we can re - arrange the formula to solve for $x$. We get $x = \frac{12}{\sin(60^{\circ})}$.
Since $\sin(60^{\circ})=\frac{\sqrt{3}}{2}\approx0.866$, then $x=\frac{12}{\frac{\sqrt{3}}{2}}=\frac{24}{\sqrt{3}}=\frac{24\sqrt{3}}{3} = 8\sqrt{3}\approx8\times1.732 = 13.9$ (rounded to the nearest tenth).
Alternatively, we can use the fact that in a 30 - 60 - 90 triangle, the sides are in the ratio $1:\sqrt{3}:2$, where the side opposite $30^{\circ}$ is the shortest side (let's call it $a$), the side opposite $60^{\circ}$ is $a\sqrt{3}$, and the hypotenuse is $2a$. The side opposite $60^{\circ}$ is 12, so $a\sqrt{3}=12$, then $a=\frac{12}{\sqrt{3}} = 4\sqrt{3}$. The hypotenuse $x = 2a=8\sqrt{3}\approx13.9$.

Answer:

13.9