# Tricky Right Triangle Problem (Picture Included)

• Oct 12th 2012, 01:06 PM
JacobFischer
Tricky Right Triangle Problem (Picture Included)
Attachment 25195
It states that Segments AD, DF, FE, EC, and CB are all equal. Triangle ACB is a Right Triangle. That is all the information given. What is Angle A?
An explanation would be great because I can't seem to understand it!
• Oct 12th 2012, 02:52 PM
MaxJasper
Re: Tricky Right Triangle Problem (Picture Included)
Try A=40.6463194 (unchecked(Lipssealed))
• Oct 12th 2012, 03:49 PM
Soroban
Re: Tricky Right Triangle Problem (Picture Included)
Hello, JacobFischer!

Quote:

Attachment 25195

Segments AD, DF, FE, EC, and CB are all equal.
Triangle ACB is a Right Triangle.
What is angle A?

Let $\displaystyle \theta$ = angle A.
We will calculate all the angles in the diagram.
Write these figures in your diagram as we go along.

Since $\displaystyle \Delta ADF$ is isosceles, $\displaystyle \angle DFA = \theta.$

Since $\displaystyle \angle EDF$ is an exterior angle, $\displaystyle \angle EDF = 2\theta.$

Since $\displaystyle \Delta EDF$ is isosceles, $\displaystyle \angle DEF = 2\theta.$

Then $\displaystyle \angle DFE \:=\:180^o - 2\theta - 2\theta \:=\:180^o - 4\theta$

Hence, $\displaystyle \angle EFC \:=\:180^o - \theta - (180^o-\theta)\:=\:3\theta$

Since $\displaystyle \Delta FEC$ is isosceles, $\displaystyle \angle ECF = 3\theta.$

Then $\displaystyle \angle FEC \:=\:180^o - 3\theta - 3\theta \:=\:180^o - 6\theta$

Hence, $\displaystyle \angle BEC \:=\:180^o - 3\theta - (180^o - 6\theta) \:=\:4\theta$

Since $\displaystyle \Delta ECB$ is isosceles, $\displaystyle \angle EBC = 4\theta$

Hence, $\displaystyle \angle ECB \:=\:180^o - 4\theta - 4\theta \:=\:180^o - 8\theta$ .[1]

But we know that: $\displaystyle \angle ECB + \angle ECF \:=\:90^o$
. . That is: .$\displaystyle \angle ECB + 3\theta \:=\:90^o$ .[2]

Substitute [1] into [2]: .$\displaystyle (180^o - 8\theta) + 3\theta \:=\:90^o$

. . and we have: .$\displaystyle \text{-}5\theta \:=\:\text{-}90^o \quad\Rightarrow\quad \boxed{\theta \:=\:18^o}$