Answer:
The length of b is \(\sqrt{2x^2(1-cos\beta)}\).
Step by step explanation:
Given information: In ΔABC, ∠A =∠C, ∠B = ß (where ß is an acute angle), and BC = x.
Since two angles are same therefore triangle ABC is an isosceles triangle and side AB and BC are congruent.
\(AB=BC=x\)
According to Law of cosine
\(b^2=a^2+c^2-2ac\cos B\)
\(b^2=x^2+x^2-2(x)(x)\cos \beta\)
\(b^2=2x^2-2x^2\cos \beta\)
\(b^2=2x^2(1-\cos \beta)\)
\(\sqrt{2x^2(1-cos\beta)}\)
Therefore the length of b is \(\sqrt{2x^2(1-cos\beta)}\).
