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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.


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up vote 0 down vote accepted

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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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

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