Question

1. find the value of x. enter your answer.

Explanation:

Step1: Identify triangle type

This is a right - triangle with one angle $60^{\circ}$, so it's a 30 - 60 - 90 triangle. 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 $2a$.

In triangle $DEF$, $\angle D = 60^{\circ}$, $\angle E=90^{\circ}$, so $\angle F = 30^{\circ}$. The hypotenuse $DF = 18$. The side $x$ is opposite $\angle D = 60^{\circ}$? Wait, no. Wait, in a right - triangle, the side opposite $30^{\circ}$ is half the hypotenuse. Wait, $\angle F=30^{\circ}$, so the side opposite $\angle F$ (which is $DE$) would be half the hypotenuse? Wait, no, let's re - label. Let's say in right - triangle $DEF$, right - angled at $E$. So $\angle E = 90^{\circ}$, $\angle D=60^{\circ}$, so $\angle F = 30^{\circ}$. The hypotenuse is $DF = 18$. The side adjacent to $\angle D$ is $DE$, the side opposite to $\angle D$ is $EF=x$, and the hypotenuse is $DF$.

We know that in a right - triangle, $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$ and $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$.

We can also use the property of 30 - 60 - 90 triangles. The side opposite $30^{\circ}$ (which is $\angle F = 30^{\circ}$) is $DE$, and it should be half the hypotenuse. Wait, no, the side opposite $30^{\circ}$ is the shortest side. So if $\angle F = 30^{\circ}$, then the side opposite $\angle F$ is $DE$, so $DE=\frac{1}{2}DF$. Since $DF = 18$, $DE = 9$. Then, using Pythagoras theorem, $x=\sqrt{DF^{2}-DE^{2}}=\sqrt{18^{2}-9^{2}}=\sqrt{324 - 81}=\sqrt{243}=9\sqrt{3}\approx15.588$. Wait, but there is another way. Alternatively, since $\angle D = 60^{\circ}$, $\cos(60^{\circ})=\frac{DE}{DF}$, but we can also use $\sin(60^{\circ})=\frac{x}{DF}$. Since $\sin(60^{\circ})=\frac{\sqrt{3}}{2}$, and $DF = 18$, then $x = DF\times\sin(60^{\circ})=18\times\frac{\sqrt{3}}{2}=9\sqrt{3}\approx15.59$. Wait, but wait, maybe I made a mistake in the angle. Wait, if $\angle D = 60^{\circ}$, and the hypotenuse is $DF = 18$, then the side $x$ (which is $EF$) is opposite $\angle D$. So $\sin(\angle D)=\frac{EF}{DF}$. So $\sin(60^{\circ})=\frac{x}{18}$. Since $\sin(60^{\circ})=\frac{\sqrt{3}}{2}$, we have $x = 18\times\frac{\sqrt{3}}{2}=9\sqrt{3}\approx15.59$. But wait, there is a simpler way. Wait, in a 30 - 60 - 90 triangle, the sides are in the ratio $1:\sqrt{3}:2$. If the hypotenuse is $2a = 18$, then $a = 9$. The side opposite $60^{\circ}$ is $a\sqrt{3}=9\sqrt{3}$, and the side opposite $30^{\circ}$ is $a = 9$. So $x$, which is opposite $60^{\circ}$, is $9\sqrt{3}\approx15.59$. But wait, maybe the problem is that the side $x$ is adjacent to $60^{\circ}$? Wait, no, let's check the angle again.

Wait, the triangle is right - angled at $E$. So vertices: $D$, $E$, $F$. Right - angle at $E$. So $DE$ and $EF$ are the legs, $DF$ is the hypotenuse. $\angle D = 60^{\circ}$, so in triangle $DEF$, $\cos(\angle D)=\frac{DE}{DF}$, $\sin(\angle D)=\frac{EF}{DF}$. So $EF = DF\times\sin(60^{\circ})$. Since $DF = 18$ and $\sin(60^{\circ})=\frac{\sqrt{3}}{2}$, then $EF=x = 18\times\frac{\sqrt{3}}{2}=9\sqrt{3}\approx15.59$. But if we consider the 30 - 60 - 90 triangle ratios, the side opposite $30^{\circ}$ (which is $\angle F = 30^{\circ}$) is $DE=\frac{1}{2}DF = 9$, and the side opposite $60^{\circ}$ (which is $EF$) is $DE\times\sqrt{3}=9\sqrt{3}$, which matches.

Step2: Calculate the value of $x$

We know that in right - triangle $DEF$ (right - angled at $E$), $\angle D = 60^{\circ}$, hypotenuse $DF = 18$. We use the sine function: $\sin(\angle D)=\frac{\text{opposite}}{\text{hypotenuse}}$. The side opposite $\angle D$ is $EF=x$, and the hypotenuse is $DF = 18$.

$\sin(60^{\circ})=\frac{x}{18}$

Since $\sin(60^{\circ})=\frac{\sqrt{3}}{2}$, we can substitute:

$\frac{\sqrt{3}}{2}=\frac{x}{18}$

Cross - multiply:

$x=\frac{18\times\sqrt{3}}{2}=9\sqrt{3}\approx15.59$ (or if we want an exact value, $9\sqrt{3}$, or a decimal approximation). But wait, maybe I got the angle wrong. Wait, if $\angle F = 30^{\circ}$, then the side adjacent to $\angle D$ is $DE$, and the side opposite is $EF$. Alternatively, we can use the cosine of $\angle F$. $\angle F = 30^{\circ}$, so $\cos(\angle F)=\frac{EF}{DF}$, so $EF = DF\times\cos(30^{\circ})=18\times\frac{\sqrt{3}}{2}=9\sqrt{3}$, which is the same result.

Answer:

$9\sqrt{3}$ (or approximately $15.6$)