Answer:
The length of b is [tex]\sqrt{2x^2(1-cos\beta)}[/tex].
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.
[tex]AB=BC=x[/tex]
According to Law of cosine
[tex]b^2=a^2+c^2-2ac\cos B[/tex]
[tex]b^2=x^2+x^2-2(x)(x)\cos \beta[/tex]
[tex]b^2=2x^2-2x^2\cos \beta[/tex]
[tex]b^2=2x^2(1-\cos \beta)[/tex]
[tex]\sqrt{2x^2(1-cos\beta)}[/tex]
Therefore the length of b is [tex]\sqrt{2x^2(1-cos\beta)}[/tex].
