Question

find the smallest angle of $\triangle hij$. assume that $u$ is a positive number.

Explanation:

Step1: Recall triangle angle-side relationship

In any triangle, the smallest angle is opposite the shortest side.

Step2: Identify side lengths

Side opposite $\angle H$: $6u$
Side opposite $\angle I$: $HJ$ (calculated via Pythagoras: $\sqrt{(14u)^2+(6u)^2}=\sqrt{196u^2+36u^2}=\sqrt{232u^2}=2u\sqrt{58}\approx15.23u$)
Side opposite $\angle J$: $14u$

Step3: Compare side lengths

$6u < 14u < 2u\sqrt{58}$, so the shortest side is $6u$, opposite $\angle H$.

Step4: Verify angle sum rule

$\angle I=91^\circ$, so $\angle H+\angle J=180^\circ-91^\circ=89^\circ$, meaning both $\angle H$ and $\angle J$ are acute, and $\angle H$ is smaller as it is opposite the shortest side.

Answer:

$\angle H$