A flag pole of height '$h$' meters is on top of a hemispherical dome of radius '$r$' meters. A man standing $7 \; m$ away from the dome sees the top of the pole at an angle of $45^\circ$. On moving $5 \; m$ away from the dome, he sees the bottom of the pole at an angle of $30^\circ$. Find the height of the pole and the radius of the dome. [Given: $\sqrt{3} = 1.732$]
$OO' = OA = r = $ Radius of the dome
$O'P = $ flag pole of height $h$ meter
$B = $ Initial point of observation such that $AB = 7 \; m$
$C = $ Final point of observation such that $BC = 5 \; m$
In $\triangle OO'C$, $\;$ $\dfrac{OO'}{OC} = \tan 30^\circ = \dfrac{1}{\sqrt{3}}$
$\implies$ $OO' = \dfrac{OC}{\sqrt{3}}$
i.e. $\;$ $OO' = \dfrac{OA + AB + BC}{\sqrt{3}} = \dfrac{OA + 12}{\sqrt{3}}$
i.e. $\;$ $OO' \times \sqrt{3} = OA + 12$
i.e. $\;$ $OO' \times \sqrt{3} = OO' + 12$
i.e. $\;$ $\left(\sqrt{3} - 1\right)OO' = 12$
$\implies$ $OO' = \dfrac{12}{0.732} = 16.39$
In $\triangle OPB$, $\;$ $\dfrac{OP}{OB} = \tan 45^\circ = 1$
$\implies$ $OP = OB$
i.e. $\;$ $h + r = r + AB$
i.e. $\;$ $h = AB = 7$
$\therefore \;$ Height of pole $= 7 \; m$
Radius of dome $= 16.39 \; m$