Solve this SAT math problem to crush the SATs this year #digitalsat

[ananyamettu]

i feel like this is a lot simpler if you just think about it as similar to the unit circle only with a radius of 2. the point sqrt3, 1 corresponds to the point sqrt3 over 2, 1/2 on the actual unit circle. the angle measure for that point would remain the same & you know that angle must be pi/6 based on the unit circle

[Jesus_IsGood-Christ]

Pythagorean theorem could have been much easier

[Metheglyn]

Radius |OA|=√(√3²+1²)=√4=2
If you mirror the radius around the x-axis and connect A and A', OAA' forms an equilateral triangle (because |AB|=|BA'|=1 => |ABA'|=2), with its three angles each being π/3 (=180°).
Now, AOB is the half of that, or (π
/3)/2=π
/6, so a=6