Answer: The answer is 3.5 inches.
Step-by-step explanation: As given in the question and shown in the attached figure that ΔAB is right angled at ∠B = 90° and let [tex]\cos \angle ACB=\dfrac{1}{2}.[/tex]. Also, hypotenuse h = 7 inches. We are to find the length of the shortest side of ΔABC.
We have,
[tex]\cos \angle ACB=\dfrac{BC}{AC}\\\\\\\Rightarrow \dfrac{1}{2}=\dfrac{b}{h}\\\\\\\Rightarrow \dfrac{1}{2}=\dfrac{b}{7}\\\\\\\Rightarrow b=\dfrac{7}{2}=3.5.[/tex]
Now, using the Pythagoras theorem, we have
[tex]b^2+p^2=h^2\\\\\Rightarrow p^2=h^2-b^2=7^2-\left(\dfrac{7}{2}\right)^2\\\\\\\Rightarrow p^2=\dfrac{3 \times 49}{4}\\\\\\\Rightarrow p=\sqrt 3\times\dfrac{7}{2}=\sqrt 3\times 3.5.[/tex]
since, √3 × 3.5 > 3.5, so, the shortest side of the triangle will be 3.5 inches.
Thus, the answer is 3.5 inches.
