A vertical tower $\text{OP}$ stands at the centre $\text{O}$ of a square $\text{ABCD}.$ Let $h$ and $b$ denote the lengths $\text{OP}$ and $\text{AB}$ respectively. Suppose $\measuredangle \text{APB} = 60^{\circ}$. Then the relationship between $h$ and $b$ can be expressed as
- $2b^2 = h^2$
- $2h^2 = b^2$
- $3b^2 = 2h^2$
- $3h^2 = 2b^2$