# Thread: Triangle in a cyrcle

1. ## Triangle in a cyrcle

Does anyone know how to solve this? It has no information, just the answer is given, which is 72deg.

2. ## Re: Triangle in a cyrcle

Hey birkoof.

You have an isosceles triangle, a hemi-circle and another circle [in the top].

Try writing down as many variables as possible with your picture and show us the work you have done to write them as formulas so we can give you a hand.

3. ## Re: Triangle in a cyrcle

Well, inside the triangle I have 2alpha + beta = 180deg.
Using the Angle in the Center Theorem I get a X that is two times bigger than my beta (beta = x/2)
Implementing that in 2alpha + beta = 180deg i came with the result that alpha is equal to 90deg - x/4.
That's all i can get from the information given.

4. ## Re: Triangle in a cyrcle

triangle with angles $\displaystyle \beta ,\beta ,\alpha +\beta$ says $\displaystyle \alpha +3\beta =180{}^{\circ}$

5. ## Re: Triangle in a cyrcle

Originally Posted by Idea
triangle with angles $\displaystyle \beta ,\beta ,\alpha +\beta$ says $\displaystyle \alpha +3\beta =180{}^{\circ}$
Thanks, this helped a lot. Now i just express $\displaystyle \beta$ from the first triangle, which is $\displaystyle 180{}^{\circ} - 2 \alpha$ , put in the $\displaystyle \alpha +3\beta =180{}^{\circ}$ and i get that $\displaystyle \alpha = 360{}^{\circ} /5$ which is $\displaystyle 72{}^{\circ}$ and the right answer. Thanks!
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