Question

consider the following figure.
(note that the figure is not drawn to scale.)
order the side lengths hi, ik, hj, ij, and kj from least to greatest.
\square < \square < \square < \square < \square

Explanation:

Step1: Analyze Triangle HIJ

In $\triangle HIJ$, angles are $\angle HIJ = 70^\circ$, $\angle HJI = 65^\circ$. So $\angle H = 180^\circ - 70^\circ - 65^\circ = 45^\circ$. By the Law of Sines, side opposite smaller angle is shorter. So $IJ < HI < HJ$ (since $\angle H = 45^\circ < \angle HIJ = 70^\circ < \angle HJI = 65^\circ$? Wait, no: $\angle H = 45^\circ$, $\angle HIJ = 70^\circ$, $\angle HJI = 65^\circ$. So order of angles: $\angle H < \angle HJI < \angle HIJ$. Thus sides: $IJ < HJ < HI$? Wait, no: side opposite $\angle H$ is $IJ$, side opposite $\angle HIJ$ is $HJ$, side opposite $\angle HJI$ is $HI$. So $\angle H = 45^\circ$ (opposite $IJ$), $\angle HIJ = 70^\circ$ (opposite $HJ$), $\angle HJI = 65^\circ$ (opposite $HI$). So $IJ < HJ < HI$ (since $45^\circ < 65^\circ < 70^\circ$).

Step2: Analyze Triangle IKJ

In $\triangle IKJ$, angles are $\angle KIJ = 62^\circ$, $\angle K = 67^\circ$. So $\angle IJK = 180^\circ - 62^\circ - 67^\circ = 51^\circ$. Order of angles: $\angle IJK = 51^\circ < \angle KIJ = 62^\circ < \angle K = 67^\circ$. Thus sides: $KJ < IK < IJ$ (side opposite $\angle IJK$ is $IK$, side opposite $\angle KIJ$ is $KJ$, side opposite $\angle K$ is $IJ$? Wait, no: side opposite $\angle K$ (67°) is $IJ$, side opposite $\angle KIJ$ (62°) is $KJ$, side opposite $\angle IJK$ (51°) is $IK$. So $\angle IJK = 51^\circ$ (opposite $IK$), $\angle KIJ = 62^\circ$ (opposite $KJ$), $\angle K = 67^\circ$ (opposite $IJ$). So $IK < KJ < IJ$ (since $51^\circ < 62^\circ < 67^\circ$). Wait, no: smaller angle → shorter side. So $\angle IJK = 51^\circ$ (smallest angle) → opposite side $IK$ is shortest. Then $\angle KIJ = 62^\circ$ → opposite side $KJ$. Then $\angle K = 67^\circ$ → opposite side $IJ$. So $IK < KJ < IJ$.

Step3: Combine Results

From $\triangle HIJ$: $IJ < HJ < HI$? Wait, no, earlier mistake. Wait in $\triangle HIJ$: angles are $\angle H = 45^\circ$ (opposite $IJ$), $\angle HIJ = 70^\circ$ (opposite $HJ$), $\angle HJI = 65^\circ$ (opposite $HI$). So angles: $45^\circ$ (∠H) < 65° (∠HJI) < 70° (∠HIJ). So sides: $IJ$ (opposite ∠H) < $HI$ (opposite ∠HJI) < $HJ$ (opposite ∠HIJ)? Wait, no: side opposite larger angle is longer. So ∠H = 45° (smallest angle) → opposite side $IJ$ (shortest in $\triangle HIJ$). ∠HJI = 65° → opposite side $HI$. ∠HIJ = 70° → opposite side $HJ$. So $IJ < HI < HJ$.

From $\triangle IKJ$: angles ∠KIJ = 62°, ∠K = 67°, ∠IJK = 51°. So ∠IJK = 51° (smallest) → opposite side $IK$ (shortest in $\triangle IKJ$). ∠KIJ = 62° → opposite side $KJ$. ∠K = 67° → opposite side $IJ$. So $IK < KJ < IJ$.

Now, compare $IJ$ from both triangles: it's the same side. So from $\triangle IKJ$, $IK < KJ < IJ$, and from $\triangle HIJ$, $IJ < HI < HJ$. So combining: $IK < KJ < IJ < HI < HJ$.

Answer:

$IK < KJ < IJ < HI < HJ$