How to find the included angle between the given known sides

Question

Question: In a triangle $ABC$, $AB = a-b$ & $BC = 2\sqrt{ab}$ , find $\angle B$ ??

Answer 1

Clue:

In this triangle ABC.If you consider conventional naming styles of sides. The length Side AC = b units , AB = c units , BC = a units.

Now we know that $a = 2\sqrt{ab}$ $\Rightarrow a^2 = 4ab$. We will get two values of a from this viz. $a = 0 $ & $a = 4b$.We cannot use a = 0 as triangle cannot exist .Use a = 4b.Find c in terms of b and substitute below.

Now use cosine rule:

$$cosB = \frac{a^2+c^2-b^2}{2ac}$$