# let ABC be a right triangle at A such that $BC=2AB$. Find $\angle ACB$

So let $\triangle ABC$ be a right triangle at vertex $A$ such that $BC=2AB$. Find the $\angle ACB$

How can I find that angle without using cosine, sine and other things?

Since I've already figure out how to find it using cos: here's my approach:

We denote $\angle ABC$ as $\alpha$ so $\cos\alpha=\frac{AB}{BC}=\frac{AB}{2AB}=\frac{1}{2}$

We do $\cos^{-1}$ to find $\angle ABC$ then we do $90^\circ-\angle ABC$ to find $\angle ACB$.

So I'm looking for alternative way.

Thanks!

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