A hard question of finding angle

**3ASV444** · October 3rd 2009, 12:32 PM

could any smart person help solve this question?
thank you so much.....

Find x where AB=BC

**pflo** · October 4th 2009, 06:26 AM

This is not a hard question, it is an impossible one. Not enough information here. Now I could be missing something, but I don't see how you can lock in an angle for x. If m<DAC = 0, then x=90. But if m<DAC = 30, then x = 60. And if m<DAC = 45, then x = 45. This is true regardless of what is going on with triangleABC. We need some other constraint on the problem.

**aidan** · October 5th 2009, 01:59 AM

Originally Posted by 3ASV444

could any smart person help solve this question?
thank you so much.....

Find x where AB=BC

$\angle x = 90 - \angle EAF$
$\angle x = \angle AEF$

**Plato** · October 5th 2009, 03:23 AM

It is a messy problem, but not impossible.
To simplify notation, let $m\left( {\angle DAC} \right) = y\;\& \,m\left( {\angle BAC} \right) = m\left( {\angle BCA} \right) = z$
The sum of measures of angles of a quadrilateral is [ $360^o$ .
So $x+y+2z+205=360$ . But $x+y=90$ , so $2z+295=360$ .
Now you can finish.

**pflo** · October 5th 2009, 08:24 PM

But even if you can solve for z, you still have 2 variables and a single equation (or two equivelant equations anyway). Even if you know z, all you're left with is x+y=90.

Isn't it? How would you finish from there?

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