Question

in △def, what is the length of segment df? image of right triangle def with right angle at e, angle at f is 60°, de = 63, options: 126, 63√2, 42√3, 21√3

Explanation:

Step1: Identify triangle type and angle

$\triangle DEF$ is right - angled at $E$. $\angle F = 60^{\circ}$, so $\angle D=30^{\circ}$. In a $30 - 60 - 90$ right triangle, the side opposite $30^{\circ}$ is half the hypotenuse. Here, $DE = 63$ is opposite $\angle F = 60^{\circ}$? Wait, no. Wait, $\angle E = 90^{\circ}$, $\angle F = 60^{\circ}$, so $\angle D=30^{\circ}$. The side opposite $\angle D$ (which is $30^{\circ}$) is $EF$, and the hypotenuse is $DF$. Wait, no, let's correct. In right - triangle $DEF$, right - angled at $E$, so $DE$ and $EF$ are legs, $DF$ is hypotenuse. $\angle F = 60^{\circ}$, so $\cos(60^{\circ})=\frac{EF}{DF}$? No, wait, $\cos(\angle F)=\frac{EF}{DF}$, but we know that in a $30 - 60 - 90$ triangle, the side opposite $30^{\circ}$ is the shorter leg. Wait, $\angle D = 180^{\circ}-90^{\circ}-60^{\circ}=30^{\circ}$. So the side opposite $\angle D$ (which is $30^{\circ}$) is $EF$, and the hypotenuse is $DF$. Wait, no, $DE$ is adjacent to $\angle D$? Wait, no, let's use trigonometric ratios. $\sin(\angle F)=\frac{DE}{DF}$. Since $\angle F = 60^{\circ}$, $DE = 63$, and $\sin(60^{\circ})=\frac{\sqrt{3}}{2}$? No, wait, $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$. For $\angle F$, the opposite side is $DE$, and the hypotenuse is $DF$. So $\sin(60^{\circ})=\frac{DE}{DF}$. But we can also use the property of $30 - 60 - 90$ triangle: the side opposite $30^{\circ}$ is half the hypotenuse. Wait, $\angle D = 30^{\circ}$, so the side opposite $\angle D$ is $EF$, and the hypotenuse is $DF$, so $EF=\frac{1}{2}DF$. But we can also use $\cos(30^{\circ})=\frac{DE}{DF}$. Wait, $\angle D = 30^{\circ}$, adjacent side to $\angle D$ is $DE$, hypotenuse is $DF$. So $\cos(30^{\circ})=\frac{DE}{DF}$. But $\cos(30^{\circ})=\frac{\sqrt{3}}{2}$, no, wait, $\cos(30^{\circ})=\frac{\sqrt{3}}{2}$, but if $\angle D = 30^{\circ}$, then $\cos(\angle D)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{DE}{DF}$. Wait, maybe I made a mistake. Let's start over.

In right - triangle $DEF$, $\angle E = 90^{\circ}$, $\angle F = 60^{\circ}$, so $\angle D=30^{\circ}$. The side opposite $30^{\circ}$ is the shortest side. The side opposite $\angle D$ (30°) is $EF$, the side opposite $\angle F$ (60°) is $DE = 63$, and the hypotenuse is $DF$. In a $30 - 60 - 90$ triangle, the ratio of sides is $1:\sqrt{3}:2$, where the side opposite $30^{\circ}$ is $x$, opposite $60^{\circ}$ is $x\sqrt{3}$, and hypotenuse is $2x$. Here, the side opposite $60^{\circ}$ (which is $DE$) is $x\sqrt{3}=63$. We need to find the hypotenuse $2x$. First, solve for $x$: $x\sqrt{3}=63\Rightarrow x = \frac{63}{\sqrt{3}}=\frac{63\sqrt{3}}{3}=21\sqrt{3}$. Then the hypotenuse $DF = 2x=42\sqrt{3}$? Wait, no, wait. Wait, if the side opposite $30^{\circ}$ is $x$, side opposite $60^{\circ}$ is $x\sqrt{3}$, hypotenuse $2x$. So if $DE$ (opposite $60^{\circ}$) is $x\sqrt{3}=63$, then $x = \frac{63}{\sqrt{3}} = 21\sqrt{3}$? No, that's not right. Wait, $\frac{63}{\sqrt{3}}=\frac{63\sqrt{3}}{3}=21\sqrt{3}$? No, $63\div\sqrt{3}=21\sqrt{3}$? Wait, $21\sqrt{3}\times\sqrt{3}=21\times3 = 63$. Yes. Then the hypotenuse $DF = 2x$? Wait, no, the hypotenuse is opposite the right angle, so hypotenuse is $DF$. The side opposite $30^{\circ}$ ($\angle D$) is $EF$, so $EF=x$, $DE = x\sqrt{3}=63$, $DF = 2x$. So from $x\sqrt{3}=63$, we get $x=\frac{63}{\sqrt{3}} = 21\sqrt{3}$? No, that can't be, because then $DF = 2x=42\sqrt{3}$. Wait, let's check with trigonometry. $\sin(60^{\circ})=\frac{DE}{DF}\Rightarrow DF=\frac{DE}{\sin(60^{\circ})}$. Since $\sin(60^{\circ})=\frac{\sqrt{3}}{2}$, then $DF=\frac{63}{\frac{\sqrt{3}}{2}}=\frac{126}{\sqrt{3}}=\frac{126\sqrt{3}}{3}=42\sqrt{3}$. Yes, that's correct.

Step2: Calculate DF

We know that in right - triangle $DEF$, $\angle E = 90^{\circ}$, $\angle F = 60^{\circ}$, $DE = 63$. Using the sine function: $\sin(\angle F)=\frac{DE}{DF}$. $\sin(60^{\circ})=\frac{\sqrt{3}}{2}$, so $\frac{\sqrt{3}}{2}=\frac{63}{DF}$. Cross - multiplying gives $DF\times\sqrt{3}=63\times2 = 126$. Then $DF=\frac{126}{\sqrt{3}}$. Rationalizing the denominator: $DF=\frac{126\sqrt{3}}{3}=42\sqrt{3}$.

Answer:

$42\sqrt{3}$