Question

consider the following figure.
(note that the figure is not drawn to scale.)
order the side lengths gh, hj, gi, gj, and ji from least to greatest.
□ < □ < □ < □ < □

Explanation:

Step1: Find angles in triangle GHJ

In $\triangle GHJ$, sum of angles is $180^\circ$. So $\angle GJH = 180^\circ - 72^\circ - 67^\circ = 41^\circ$. By the Law of Sines, side opposite larger angle is longer. So in $\triangle GHJ$: $\angle H = 72^\circ$, $\angle G = 67^\circ$, $\angle J = 41^\circ$. Thus, $GJ < GH < HJ$ (since opposite angles: $HJ$ opposite $67^\circ$, $GH$ opposite $41^\circ$? Wait, no: side opposite $\angle H$ is $GJ$, opposite $\angle G$ is $HJ$, opposite $\angle J$ is $GH$. So $\angle J = 41^\circ$ (smallest), $\angle G = 67^\circ$, $\angle H = 72^\circ$ (largest). So $GH < GJ < HJ$ (since side opposite smaller angle is shorter: $GH$ opposite $41^\circ$, $GJ$ opposite $72^\circ$? Wait, no, correction: $\angle H = 72^\circ$ (opposite $GJ$), $\angle G = 67^\circ$ (opposite $HJ$), $\angle J = 41^\circ$ (opposite $GH$). So side lengths: $GH$ (opposite $41^\circ$) < $GJ$ (opposite $67^\circ$) < $HJ$ (opposite $72^\circ$).

Step2: Find angle in triangle GIJ

In $\triangle GIJ$, $\angle GJI = 180^\circ - 41^\circ = 139^\circ$ (linear pair with $\angle GJH$). Then $\angle G = 180^\circ - 139^\circ - 12^\circ = 29^\circ$? Wait, no, $\triangle GIJ$: angles at $J$ is $180 - 41 = 139^\circ$? No, $\angle GJI$ is supplementary to $\angle GJH$ (since $H$, $J$, $I$ are colinear). So $\angle GJI = 180^\circ - 41^\circ = 139^\circ$. Then in $\triangle GIJ$, angles: $\angle I = 12^\circ$, $\angle J = 139^\circ$, so $\angle G = 180 - 139 - 12 = 29^\circ$. Now, side opposite $\angle I = 12^\circ$ is $GJ$, opposite $\angle G = 29^\circ$ is $JI$, opposite $\angle J = 139^\circ$ is $GI$. So in $\triangle GIJ$: $GJ < JI < GI$ (since $\angle I = 12^\circ$ (smallest), $\angle G = 29^\circ$, $\angle J = 139^\circ$ (largest)).

Step3: Analyze $\triangle GIJ$ and $\triangle GHJ$ together

From $\triangle GHJ$: $GH < GJ < HJ$. From $\triangle GIJ$: $GJ < JI < GI$. Now, compare $JI$ with $GH$, $GJ$, $HJ$. We know $GJ < JI$ (from $\triangle GIJ$) and $GH < GJ$ (from $\triangle GHJ$), so $GH < GJ < JI$. Now compare $JI$ with $HJ$: in $\triangle GHJ$, $HJ$ is opposite $67^\circ$, in $\triangle GIJ$, $JI$ is opposite $29^\circ$? Wait, no, earlier mistake. Let's re-express all angles:

In $\triangle GHJ$: $\angle H = 72^\circ$, $\angle G = 67^\circ$, $\angle J = 41^\circ$. So sides: $GH$ (opposite $\angle J = 41^\circ$) < $GJ$ (opposite $\angle H = 72^\circ$) < $HJ$ (opposite $\angle G = 67^\circ$)? Wait, no, Law of Sines: $\frac{GH}{\sin \angle J} = \frac{GJ}{\sin \angle H} = \frac{HJ}{\sin \angle G}$. So $\angle J = 41^\circ$, $\angle H = 72^\circ$, $\angle G = 67^\circ$. So $\sin 41^\circ < \sin 67^\circ < \sin 72^\circ$ (since $\sin$ increases in $[0, 90^\circ]$). Thus, $GH < HJ < GJ$? Wait, no: $GH$ is opposite $\angle J = 41^\circ$, so $GH = k \sin 41^\circ$; $GJ$ is opposite $\angle H = 72^\circ$, so $GJ = k \sin 72^\circ$; $HJ$ is opposite $\angle G = 67^\circ$, so $HJ = k \sin 67^\circ$. Since $\sin 41^\circ \approx 0.656$, $\sin 67^\circ \approx 0.920$, $\sin 72^\circ \approx 0.951$. So $GH < HJ < GJ$.

In $\triangle GIJ$: $\angle I = 12^\circ$, $\angle GJI = 180^\circ - 41^\circ = 139^\circ$, so $\angle G = 180 - 12 - 139 = 29^\circ$. So sides: $JI$ (opposite $\angle G = 29^\circ$), $GJ$ (opposite $\angle I = 12^\circ$), $GI$ (opposite $\angle J = 139^\circ$). By Law of Sines: $\frac{GJ}{\sin 12^\circ} = \frac{JI}{\sin 29^\circ} = \frac{GI}{\sin 139^\circ}$. $\sin 12^\circ \approx 0.208$, $\sin 29^\circ \approx 0.485$, $\sin 139^\circ = \sin 41^\circ \approx 0.656$. So $GJ < JI < GI$ (since $GJ$ is opposite $12^\circ$ (smallest), $JI$ opposite $29^\circ$, $GI$ opposite $139^\circ$ (largest)).

Now, combine both triangles:

From $\triangle GHJ$: $GH < HJ < GJ$ (wait, no, earlier calculation with sines: $GH = k \sin 41^\circ \approx 0.656k$, $HJ = k \sin 67^\circ \approx 0.920k$, $GJ = k \sin 72^\circ \approx 0.951k$. So $GH < HJ < GJ$.

From $\triangle GIJ$: $GJ \approx 0.951k$, $JI = m \sin 29^\circ \approx 0.485m$, $GI = m \sin 139^\circ \approx 0.656m$. Wait, but $GJ$ is common? Wait, no, $GJ$ is a side in both triangles? Wait, the figure: $H-J-I$ are colinear, $G$ is a vertex. So $GJ$ is a segment from $G$ to $J$, $GH$ from $G$ to $H$, $HJ$ from $H$ to $J$, $JI$ from $J$ to $I$, $GI$ from $G$ to $I$.

So in $\triangle GHJ$: angles at $H=72^\circ$, $G=67^\circ$, $J=41^\circ$. So sides: $GH$ (opposite $J=41^\circ$), $GJ$ (opposite $H=72^\circ$), $HJ$ (opposite $G=67^\circ$). So $GH < HJ < GJ$ (since $41^\circ < 67^\circ < 72^\circ$, so side opposite smaller angle is shorter: $GH$ opposite $41^\circ$, $HJ$ opposite $67^\circ$, $GJ$ opposite $72^\circ$).

In $\triangle GIJ$: angles at $I=12^\circ$, $J=180-41=139^\circ$, $G=29^\circ$. So sides: $GJ$ (opposite $I=12^\circ$), $JI$ (opposite $G=29^\circ$), $GI$ (opposite $J=139^\circ$). So $GJ < JI < GI$ (since $12^\circ < 29^\circ < 139^\circ$, so side opposite smaller angle: $GJ$ opposite $12^\circ$, $JI$ opposite $29^\circ$, $GI$ opposite $139^\circ$).

Now, combine the two:

From $\triangle GHJ$: $GH < HJ < GJ$

From $\triangle GIJ$: $GJ < JI < GI$

So putting together: $GH < HJ < GJ < JI < GI$? Wait, no, wait: $GJ$ is in both. From $\triangle GHJ$, $HJ < GJ$; from $\triangle GIJ$, $GJ < JI$. So $HJ < GJ < JI$? But earlier in $\triangle GHJ$, $GH < HJ$. So $GH < HJ < GJ < JI < GI$? Wait, let's check the angles again.

Wait, in $\triangle GHJ$: $\angle H = 72^\circ$, $\angle G = 67^\circ$, $\angle J = 41^\circ$. So side opposite $\angle H$ (72°) is $GJ$, opposite $\angle G$ (67°) is $HJ$, opposite $\angle J$ (41°) is $GH$. So $GH$ (opposite 41°) < $HJ$ (opposite 67°) < $GJ$ (opposite 72°). Correct, because 41 < 67 < 72, so side opposite smaller angle is shorter.

In $\triangle GIJ$: $\angle I = 12^\circ$, $\angle GJI = 180 - 41 = 139^\circ$, $\angle G = 180 - 12 - 139 = 29^\circ$. So side opposite $\angle I$ (12°) is $GJ$, opposite $\angle G$ (29°) is $JI$, opposite $\angle J$ (139°) is $GI$. So $GJ$ (opposite 12°) < $JI$ (opposite 29°) < $GI$ (opposite 139°). Correct, since 12 < 29 < 139.

Now, $GJ$ is in both. From $\triangle GHJ$, $HJ < GJ$; from $\triangle GIJ$, $GJ < JI$. So $HJ < GJ < JI$. And from $\triangle GHJ$, $GH < HJ$. So combining: $GH < HJ < GJ < JI < GI$.

Let's verify with approximate lengths. Let’s assume $GH = 1$ (opposite 41°). Then by Law of Sines in $\triangle GHJ$: $\frac{GH}{\sin 41^\circ} = \frac{HJ}{\sin 67^\circ} = \frac{GJ}{\sin 72^\circ}$. So $HJ = \frac{\sin 67^\circ}{\sin 41^\circ} \approx \frac{0.920}{0.656} \approx 1.40$. $GJ = \frac{\sin 72^\circ}{\sin 41^\circ} \approx \frac{0.951}{0.656} \approx 1.45$.

In $\triangle GIJ$, $\frac{GJ}{\sin 12^\circ} = \frac{JI}{\sin 29^\circ} = \frac{GI}{\sin 139^\circ}$. $GJ \approx 1.45$, so $JI = \frac{\sin 29^\circ}{\sin 12^\circ} \times GJ \approx \frac{0.485}{0.208} \times 1.45 \approx 2.33 \times 1.45 \approx 3.38$. $GI = \frac{\sin 139^\circ}{\sin 12^\circ} \times GJ \approx \frac{0.656}{0.208} \times 1.45 \approx 3.15 \times 1.45 \approx 4.57$. Wait, but $\sin 139^\circ = \sin 41^\circ \approx 0.656$, so $GI = \frac{\sin 41^\circ}{\sin 12^\circ} \times GJ \approx \frac{0.656}{0.208} \times 1.45 \approx 3.15 \times 1.45 \approx 4.57$. And $JI = \frac{\sin 29^\circ}{\sin 12^\circ} \times GJ \approx 2.33 \times 1.45 \approx 3.38$. So $GH=1$, $HJ\approx1.40$, $GJ\approx1.45$, $JI\approx3.38$, $GI\approx4.57$. So order: $GH < HJ < GJ < JI < GI$.

Answer:

$GH < HJ < GJ < JI < GI$