• Apr 23rd 2008, 05:36 PM
stellina_91
Find a in the diagram below, giving all reasons.

Thanks.
• Apr 23rd 2008, 10:02 PM
Soroban
Hello, stellina_91!

I assume that $EFHD$ is a straight line.

In $\Delta CFD\!:\;\angle EFI$ is an exterior angle.
. . Hence: . $\angle EFI \:=\:\angle C + \angle D \:=\:20^o + 35^o \quad\Rightarrow\quad \angle EFI\:=\:55^o$

In $\Delta AIB\!:\;\angle EIF$ is an exterior angle.
. . Hence: . $\angle EIF \:=\:\angle A + \angle B \:=\:45^o + 30^o\quad\Rightarrow\quad\angle EIF = 75^o$

In $\Delta IEF\!:\;a \:=\:\angle IEF \:=\:180^o - 55^o - 75^o \quad\Rightarrow\quad\boxed{ a\;=\;50^o}$

• Apr 24th 2008, 10:25 AM
gdbaillargeon
Quote:

Originally Posted by stellina_91
Find a in the diagram below, giving all reasons.

Thanks.

The first thing is to realize that in each triangle shown all the angles will add up to 180 degrees. Start there and remember the adjacent opposite angles are equal