A statue $1.6 \; m$ tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is $60^\circ$ and from the same point the angle of elevation of the top of the pedestal is $40^\circ$. Find the height of the pedestal. [Given: $\tan 40^\circ = 0.8391, \; \sqrt{3} = 1.732$]
$PS =$ Statue of height $1.6 \; m$
$OP =$ Pedestal
$A =$ Point of observation
In $\triangle SOA$, $\;$ $\dfrac{SO}{OA} = \tan 60^\circ = \sqrt{3}$
$\implies$ $OA = \dfrac{SO}{\sqrt{3}} = \dfrac{SP + PO}{\sqrt{3}}$
i.e. $\;$ $OA = \dfrac{1.6 + PO}{1.732}$ $\;\;\; \cdots \; (1)$
In $\triangle POA$, $\;$ $\dfrac{PO}{OA} = \tan 40^\circ = 0.8391$
$\implies$ $OA = \dfrac{PO}{0.8391}$ $\;\;\; \cdots \; (2)$
$\therefore \;$ We have from equations $(1)$ and $(2)$,
$\dfrac{1.6 + PO}{1.732} = \dfrac{PO}{0.8391}$
i.e. $\;$ $1.34256 + 0.8391 \; PO = 1.732 \; PO$
i.e. $\;$ $0.8929 \; PO = 1.34256$ $\implies$ $PO = 1.5036$
$\therefore \;$ Height of the pedestal $= 1.5 \; m$