∠C = 45° or 30°? I tried to solve the following math problem using both 'Pythagorean Theorem' and 'Trigonometric Ratios' and I got different answers. No matter what method is followed, both answer should be the same. I don't understand what I'm doing wrong.
Q. In △ABC, ∠B = 90°. If AC = 2AB, then what is the value of ∠C?
I'm not allowed to add any images for some reason, please click here to see the triangle.
'Pythagorean Theorem' approach: 

AC2 = AB2 + BC2
  2AB2 = AB2 + BC2 [since, AC = 2AB]
  2AB2 -  AB2 = BC2
  AB2 = BC2
  AB = BC 
  If AB = BC, then △ABC is an isosceles right triangle and ∠C = ∠B = 45°.

'Trigonometric Ratios' approach:

Suppose, ∠C = θ
  sin θ = AB/AC
  sin θ = AB/2AB [since, AC = 2AB]
  sin θ = 1/2
  sin θ = sin 30°
θ = 30°
  So, ∠C = 30°

 A: By Pythagora's theorem, we have: $$\overline{BC}=\sqrt{4\overline{AB}^2-\overline{AB}^2}=\sqrt{3\overline{AB}^2}\sqrt{3}\overline{AB}$$
Now we can caculate $\tan(C)$, so we have:
$$\tan(C)=\frac{\overline{AB}}{\overline{BC}}=\frac{1}{\sqrt{3}}$$
From here we deduce that $C=\pi/6$ and $A=\pi/3$.
Note that in Pythagora approach your solution is not correct because you have to say $\overline{BC}^2=4\overline{AB}^2$ not $\overline{BC}=2\overline{BC}^2$
A: It is a good practice to enclose the operand within brackets when operation is done on two or more terms.
$$ (2 AB)^2= 4 AB^2 \ne 2 AB^2 $$
Except this error all else you worked out is correct, and when actually corrected you get $ \angle A =60^{\circ},\angle B=30^{\circ}$ in both ways.
A: When you substituted $AC=2AB$ into $AC^2 = AB^2 + BC^2$, the correct result is
$$
(2AB)^2 = AB^2 + BC^2
$$
So you forgot to square the factor $2$. The latter method is correct.

This is my 200th answer, by the way! Worth celebrating :)
A: We prove the correct angle with no trigonometry:
Extend $\overline {AB}$ through $B$ to point $D$ such that $CD=AB$.  Then $\angle CBD$ is supplemental to $\angle CBA$ and thus measures $90°$.  This forces $\triangle ABC$ to be congruent with $\triangle DBC$.  Therefore their corresponding sides $\overline {CA}, \overline {CD}$ are also congruent, each measuring $2AB$.  By construction $BD=2AB$ as well rendering $\triangle ABD$ equilateral, therefore also equiangular.  Putting in the sum of interior angles equal to 180° then renders the measure of each individual angle, including $\angle A$, equal to 60°.  $\angle C$ is complementary to $\angle A$ and thu measures 30°.
