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

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

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

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