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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

  1. $2b^2 = h^2$
  2. $2h^2 = b^2$
  3. $3b^2 = 2h^2$
  4. $3h^2 = 2b^2$
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