# Thread: A hard question of finding angle

1. ## A hard question of finding angle

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

Find x where AB=BC

2. 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.

3. Originally Posted by 3ASV444
could any smart person help solve this question?
thank you so much.....

Find x where AB=BC
$\angle x = 90 - \angle EAF$
$\angle x = \angle AEF$

4. 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.

5. 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?