Angle between two vectors given magnitudes only I have a physics problem about vectors:

Two vectors $\boldsymbol{A}$ and $\boldsymbol{B}$ have precisely equal magnitudes. For the magnitude of $\boldsymbol{A} + \boldsymbol{B}$ to be $100$ times larger than the magnitude of $\boldsymbol{A} - \boldsymbol{B}$, what must be the angle between them?

I have tried to draw a diagram and calculate the problem with geometrical methods with two simultaneous equations of the form $c^2 = a^2 + b^2 - 2ab \cos θ$:$$
|\boldsymbol{A} + \boldsymbol{B}|² = |\boldsymbol{A}|² + |\boldsymbol{B}|² - 2|\boldsymbol{A}||\boldsymbol{B}|\cos θ\\
|\boldsymbol{A} - \boldsymbol{B}|² = |\boldsymbol{A}|² + |\boldsymbol{B}|² - 2|\boldsymbol{A}||\boldsymbol{B}|\cos(π - θ)
$$
Equating these two equations in terms of $θ$ gives$$
\cos θ = -\frac{9999|\boldsymbol{A} + \boldsymbol{B}|²}{|4|\boldsymbol{A}|²|}.
$$
This is as far as i could get, any help solving the problem will be greatly appreciated
 A: You haven't used the fact that $|A| = |B|$
$|A+B| = 100|A-B|\\
|A|^2 + |B|^2 + 2|A||B|\cos \theta = 100^2(|A|^2 + |B|^2 - 2|A||B|\cos \theta)$
let $a = |A| = |B|$
$a^2 + a^2 + 2a^2\cos\theta = 100^2(a^2 + a^2 - 2a^2\cos\theta)$
Isolate $\cos \theta$ and simplify
$2(100^2+ 1)a^2 \cos \theta = 2a^2(100^2-1)\\
\theta = \arccos \frac {100^2 - 1}{100^2 + 1}$

A: Intuitively they need to be almost in the same direction.  The sum will be (almost) twice the length of one vector and the difference will be the (small) sideways component between them.  We can divide by the length and work with unit vectors, then choose our coordinates so that $A=(1,0),B=(\cos \theta, \sin \theta)$.  The angle between them is then $\pi-\theta$
The length of the sum is then $\sqrt{(1+\cos \theta)^2+\sin^2 \theta}=\sqrt{2+2\cos \theta}$.
The length of the difference is $\sqrt{(1-\cos \theta)^2+\sin^2 \theta}=\sqrt{2-2 \cos \theta}$
Let $\cos \theta=c$ to save typing and we want 
$$\frac {\sqrt{2+2c}}{\sqrt{2-2c}}=100\\\frac {2+2c}{2-2c}=10000\\
2+2c=20000-20000c\\c=\frac{19998}{20002}\\ \theta=\arccos \frac {19998}{20002}
\approx 0.0199999$$
