\(ABC\) is an isosceles triangle.

\(B = C = 80 ^\circ\).

\(CF\) at \(30^\circ\) to \(AC\) cuts \(AB\) in \(F.\)
\(BE\) at \(20^\circ\) to \(AB\) cuts \(AC\) in \(E\).

Find then angle \(BEF \) in degrees.

×

Problem Loading...

Note Loading...

Set Loading...