Question

find the smallest angle of △hij.

Explanation:

Step1: Recall angle - side relationship

In a triangle, the smallest angle is opposite the shortest side. The shortest side of $\triangle HIJ$ is $HI = 33$ yd. The angle opposite $HI$ is $\angle J$.

Step2: Use the Law of Cosines

The Law of Cosines is $c^{2}=a^{2}+b^{2}-2ab\cos C$. For $\angle J$, let $a = 33$, $b = 40$, and $c = 41$. Then $\cos J=\frac{33^{2}+40^{2}-41^{2}}{2\times33\times40}$.
First, calculate the numerator:
$33^{2}=33\times33 = 1089$, $40^{2}=1600$, $41^{2}=1681$.
$33^{2}+40^{2}-41^{2}=1089 + 1600-1681=1008$.
Then, calculate the denominator: $2\times33\times40 = 2640$.
So, $\cos J=\frac{1008}{2640}=\frac{21}{55}\approx0.3818$.

Step3: Find the angle

$J=\cos^{-1}(\frac{21}{55})\approx67.6^{\circ}$

Answer:

$\angle J\approx67.6^{\circ}$