Triangle in a cyrcle

Dec 2016
3
0
Austria
Does anyone know how to solve this? It has no information, just the answer is given, which is 72deg.
Screenshot_51.png
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
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.
 
Dec 2016
3
0
Austria
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.
 
Jun 2013
1,151
614
Lebanon
triangle with angles \(\displaystyle \beta ,\beta ,\alpha +\beta \) says \(\displaystyle \alpha +3\beta =180{}^{\circ}\)
 
Dec 2016
3
0
Austria
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!