Let sides of a $\triangle ABC$ are in A.P. and $a < \min\{b, c\}$, then $\cos A$ is equal to:
Step-by-Step Solution
Key Concept: Substituting linear constraints on side lengths directly into the cosine rule yields specific angle expressions.
For a triangle with sides $a, b, c$, apply the cosine rule $\cos A = \frac{b^2+c^2-a^2}{2bc}$. Case 1: when $2c=a+b$, substitute to get $\cos A = \frac{b^2+c^2-(\frac{a+b}{2})^2}{2bc} = \frac{4bc-3c^2}{4b}$. Case 2: when $2b=a+c$, similarly derive $\cos A = \frac{4bc-3b^2}{4c}$.
Correct Answer: 2,3