0 0 votes 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 $2b^2 = h^2$ $2h^2 = b^2$ $3b^2 = 2h^2$ $3h^2 = 2b^2$ Quantitative Aptitude cat2003-1 quantitative-aptitude trigonometry height-distance + – go_editor 14.2k points 1.2k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.