# 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!

## 1 Answer

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

• I don't understand those terms, could you clarify a bit more please sir :) – user138849 Mar 29 '14 at 13:40
• 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 ? – DonAntonio Mar 29 '14 at 13:42
• I'm not a native english speaker so I don't know what equilateral triangle means, bisector, median – user138849 Mar 29 '14 at 13:52
• 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. – DonAntonio Mar 29 '14 at 13:56
• ookay i understand thanks anyway – user138849 Mar 29 '14 at 13:57