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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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Hint: take an equilateral triangle and draw one of the altitudes. Remember this altitude is also an angle bisector and a median, so...

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  • $\begingroup$ I don't understand those terms, could you clarify a bit more please sir :) $\endgroup$ – user138849 Mar 29 '14 at 13:40
  • $\begingroup$ Altitude=height=segment of straight line from a vertex to the opposite side (a cevian) which is perpendicular to that side. What else isn't clear, @user138849 ? $\endgroup$ – DonAntonio Mar 29 '14 at 13:42
  • $\begingroup$ I'm not a native english speaker so I don't know what equilateral triangle means, bisector, median $\endgroup$ – user138849 Mar 29 '14 at 13:52
  • $\begingroup$ Then I can't help you, @user138849...I don't even know what your mother tongue is. If you ask in english then try to look for the corresponding terms in english. $\endgroup$ – DonAntonio Mar 29 '14 at 13:56
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    $\begingroup$ ookay i understand thanks anyway $\endgroup$ – user138849 Mar 29 '14 at 13:57

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