A hall has a square floor of dimension 10 m $$\times$$ 10 m (see the figure) and vertical walls. If the angle GPH between the diagonals AG and BH is $$\cos^{-1}\frac{1}{5}$$, then the height of the hall (in meters) is:

$$P = \left( \frac{0+10}{2}, \frac{10+0}{2}, \frac{h+0}{2} \right)$$

$$= \left( 5, 5, \frac{h}{2} \right)$$

$$\vec{PG} = G - P = (10, 0, h) - \left( 5, 5, \frac{h}{2} \right)$$

$$= \left( 5, -5, \frac{h}{2} \right)$$

$$\vec{PH} = H - P = (0, 0, h) - \left( 5, 5, \frac{h}{2} \right)$$

$$= \left( -5, -5, \frac{h}{2} \right)$$

Angle $$GPH = \theta$$, $$\cos \theta = \frac{1}{5}$$.

Use dot product rule:

$$\vec{PG} \cdot \vec{PH} = |\vec{PG}| |\vec{PH}| \cos \theta$$

$$(5)(-5) + (-5)(-5) + \frac{h}{2} \left( \frac{h}{2} \right) = \left( \sqrt{50 + \frac{h^2}{4}} \right)^2 \cdot \frac{1}{5}$$

$$-25 + 25 + \frac{h^2}{4} = \left( 50 + \frac{h^2}{4} \right) \cdot \frac{1}{5}$$

Multiply by 20:

$$5h^2 = 200 + h^2$$

$$4h^2 = 200$$

$$h^2 = 50$$

$$h = \sqrt{50} = 5 \cdot \sqrt{2}$$