Question

quiz instructions
complete during class time wednesday january 14th
question 6 15 pts
δfhx has angle measures m∠hxf = 90° and m∠hfx = 30°, and side length hx = 6√3. what are the exact measures of fx and fh?

Explanation:

Step1: Identify triangle type

$\triangle FHX$ is a right - triangle with $\angle HXF = 90^{\circ}$ and $\angle HFX=30^{\circ}$, so $\angle FHX = 60^{\circ}$. In a $30 - 60 - 90$ right - triangle, the sides are in the ratio $1:\sqrt{3}:2$, where the side opposite $30^{\circ}$ is the shortest side, the side opposite $60^{\circ}$ is $\sqrt{3}$ times the shortest side, and the hypotenuse is twice the shortest side.

Step2: Analyze sides

We know that $HX = 6\sqrt{3}$. Let's identify which angle $HX$ is opposite to. $\angle HFX = 30^{\circ}$, the side opposite $\angle HFX$ is $HX$. Wait, no, in $\triangle FHX$, $\angle HXF = 90^{\circ}$, so the sides: $HX$ is opposite $\angle HFX$ (which is $30^{\circ}$), $FH$ is opposite $\angle HXF$ (which is $90^{\circ}$)? No, wait, vertices: $F$, $H$, $X$. $\angle HXF$ is at $X$, between $H$ and $F$. So the sides: $FX$ is the hypotenuse (opposite the right angle $\angle HXF$), $FH$ is opposite $\angle HX F$? No, let's label the triangle properly. Right angle at $X$, so legs are $HX$ and $XH$? Wait, no, vertices are $F$, $H$, $X$. So sides: $FX$ (hypotenuse), $HX$ (one leg), $FH$ (another leg). $\angle HFX = 30^{\circ}$, so the side opposite $\angle HFX$ is $HX$, the side opposite $\angle FHX$ (which is $60^{\circ}$) is $FX$? Wait, no, in a right - triangle, the hypotenuse is opposite the right angle. So right angle at $X$, so hypotenuse is $FH$? Wait, I made a mistake. Let's re - label:

In $\triangle FHX$, $\angle X = 90^{\circ}$, so sides:
- Opposite $\angle F$ ($30^{\circ}$): $HX$
- Opposite $\angle H$ ($60^{\circ}$): $FX$
- Hypotenuse (opposite $\angle X$ ($90^{\circ}$)): $FH$

In a $30 - 60 - 90$ triangle, the ratio of sides is: opposite $30^{\circ}$ : opposite $60^{\circ}$ : hypotenuse $=1:\sqrt{3}:2$

We know $HX$ (opposite $30^{\circ}$) is $6\sqrt{3}$. Wait, no, if $\angle F = 30^{\circ}$, then the side opposite $\angle F$ is $HX$, the side opposite $\angle H$ (60°) is $FX$, and hypotenuse $FH$.

The ratio of opposite $30^{\circ}$ (HX) : opposite $60^{\circ}$ (FX) : hypotenuse (FH) is $1:\sqrt{3}:2$. Wait, if $HX$ (opposite $30^{\circ}$) is $a$, then $FX$ (opposite $60^{\circ}$) is $a\sqrt{3}$, and $FH$ (hypotenuse) is $2a$. But we are given $HX = 6\sqrt{3}$. Wait, that would mean $a = 6\sqrt{3}$, then $FX=a\sqrt{3}=6\sqrt{3}\times\sqrt{3}=18$, and $FH = 2a = 12\sqrt{3}$? Wait, no, maybe I mixed up the angles.

Wait, let's use trigonometry. In right - triangle $FXH$ (right - angled at $X$), $\cos(\angle HFX)=\frac{FX}{FH}$? No, $\cos(\angle HFX)=\frac{FH}{FX}$, because $\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$. $\angle HFX = 30^{\circ}$, adjacent side to $\angle HFX$ is $FH$, hypotenuse is $FX$. $\sin(\angle HFX)=\frac{HX}{FX}$, because $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$.

We know $\sin(30^{\circ})=\frac{HX}{FX}$, and $HX = 6\sqrt{3}$, $\sin(30^{\circ})=\frac{1}{2}$. So $\frac{1}{2}=\frac{6\sqrt{3}}{FX}$, then $FX = 12\sqrt{3}$? Wait, but the first box has 18. Wait, maybe I labeled the triangle wrong. Let's assume that the right angle is at $H$. Wait, the problem says $\angle HXF = 90^{\circ}$, so right angle at $X$. Let's try again.

$\angle HXF = 90^{\circ}$, $\angle HFX = 30^{\circ}$, so $\angle FHX=60^{\circ}$. So sides:

- $HX$: opposite $\angle HFX$ (30°), length $a$
- $FH$: opposite $\angle HXF$ (90°), length $2a$ (hypotenuse)
- $FX$: opposite $\angle FHX$ (60°), length $a\sqrt{3}$

Wait, but we are given $HX = 6\sqrt{3}$. If $HX$ is opposite $30^{\circ}$, then $a = 6\sqrt{3}$, so $FX=a\sqrt{3}=6\sqrt{3}\times\sqrt{3}=18$, and $FH = 2a=12\sqrt{3}$. Ah, that matches the first box (FX = 18). So:

For $FX$:
We use the ratio of the side opposite $60^{\circ}$ (FX) to the side opposite $30^{\circ}$ (HX) is $\sqrt{3}:1$. So $FX = HX\times\sqrt{3}$? Wait, no, if $HX$ is opposite $30^{\circ}$, then $FX$ (opposite $60^{\circ}$) is $HX\times\sqrt{3}$? Wait, $HX = 6\sqrt{3}$, then $FX=6\sqrt{3}\times\sqrt{3}=18$ (since $\sqrt{3}\times\sqrt{3}=3$, $6\times3 = 18$).

For $FH$:
The side opposite $90^{\circ}$ (hypotenuse) is $FH$? No, wait, right angle at $X$, so hypotenuse is $FH$? No, right angle at $X$, so hypotenuse is $FH$ (connecting $F$ and $H$), and legs are $HX$ (connecting $H$ and $X$) and $FX$ (connecting $F$ and $X$). Wait, I think I had the labels wrong. Let's correct:

Right angle at $X$, so:
- Legs: $HX$ (from $H$ to $X$) and $FX$ (from $F$ to $X$)
- Hypotenuse: $FH$ (from $F$ to $H$)

$\angle HFX = 30^{\circ}$, which is at $F$, between $H$ and $X$. So in $\triangle FHX$, angle at $F$ is $30^{\circ}$, right angle at $X$. So:

$\sin(\angle HFX)=\frac{HX}{FH}$ (opposite over hypotenuse)

$\cos(\angle HFX)=\frac{FX}{FH}$ (adjacent over hypotenuse)

We know $\angle HFX = 30^{\circ}$, $HX = 6\sqrt{3}$

Since $\sin(30^{\circ})=\frac{1}{2}=\frac{HX}{FH}$, then $FH = 2\times HX=2\times6\sqrt{3}=12\sqrt{3}$

Since $\cos(30^{\circ})=\frac{\sqrt{3}}{2}=\frac{FX}{FH}$, and $FH = 12\sqrt{3}$, then $FX=\frac{\sqrt{3}}{2}\times12\sqrt{3}=\frac{12\times3}{2}=18$

So we already have $FX = 18$, and we need to find $FH$.

Step3: Calculate FH

Using the $30 - 60 - 90$ triangle ratio, the side opposite $30^{\circ}$ ($HX$) is $6\sqrt{3}$, the hypotenuse $FH$ (opposite the right angle) is twice the side opposite $30^{\circ}$? Wait, no, in a $30 - 60 - 90$ triangle, the hypotenuse is twice the shorter leg (the leg opposite $30^{\circ}$). Wait, $HX$ is opposite $30^{\circ}$, so it is the shorter leg? But $HX = 6\sqrt{3}$, and if the shorter leg is $a$, then hypotenuse is $2a$. But $6\sqrt{3}$ is not a short leg in the ratio $1:\sqrt{3}:2$ if we consider $a$ as the short leg. Wait, no, maybe the short leg is $FH$? No, we saw from trigonometry that $FH = 12\sqrt{3}$ and $FX = 18$.

Wait, let's use the Pythagorean theorem to verify. If $FX = 18$ (one leg), $HX = 6\sqrt{3}$ (another leg), then $FH=\sqrt{FX^{2}-HX^{2}}=\sqrt{18^{2}-(6\sqrt{3})^{2}}=\sqrt{324 - 108}=\sqrt{216}=6\sqrt{6}$? No, that's wrong. Wait, I messed up the legs and hypotenuse.

Correct labeling: Right angle at $X$, so legs are $HX$ and $X F$, hypotenuse is $FH$. So:

$HX^{2}+XF^{2}=FH^{2}$

$\angle HFX = 30^{\circ}$, so $\tan(\angle HFX)=\frac{HX}{XF}$

$\tan(30^{\circ})=\frac{1}{\sqrt{3}}=\frac{6\sqrt{3}}{XF}$

Then $XF = 6\sqrt{3}\times\sqrt{3}=18$ (which matches the first box)

Then, using $\sin(\angle HFX)=\frac{HX}{FH}$

$\sin(30^{\circ})=\frac{1}{2}=\frac{6\sqrt{3}}{FH}$

So $FH = 12\sqrt{3}$

Or using the $30 - 60 - 90$ ratio: 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 leg, the side opposite $60^{\circ}$ is $\sqrt{3}$ times the shortest leg, and the hypotenuse is twice the shortest leg.

Here, $\angle HFX = 30^{\circ}$, so the side opposite it ($HX$) is the shortest leg. The side opposite $60^{\circ}$ (which is $\angle FHX$) is $XF$, and it should be $\sqrt{3}$ times the shortest leg. Wait, $HX$ is the shortest leg ($a = 6\sqrt{3}$), then $XF=a\sqrt{3}=6\sqrt{3}\times\sqrt{3}=18$ (which is correct), and the hypotenuse $FH = 2a=2\times6\sqrt{3}=12\sqrt{3}$

Answer:

For $FH$, the exact measure is $12\sqrt{3}$ (and $FX = 18$ as given in the first box). So the measure of $FH$ is $12\sqrt{3}$.