The angle of elevation of the top of a vertical tower from a point on the ground is $60^{\circ}$. From another point 10m vertically above the first, its angle of elevation is $45^{\circ}$. Find the height of tower.
OT: Tower of height 'h'
B: Point 10m vertically above the first point A
From the figure, $OT=h$, $PT=h-10$, $BA=10$, $OA=PB$
From triangle TOA, $\dfrac{OT}{OA} = \tan 60^{\circ} = \sqrt{3}$
$\implies$ $OA=\dfrac{h}{\sqrt{3}}$ $\cdots$ (1)
From triangle TPB, $\dfrac{PT}{PB} = \tan 45^{\circ} = 1$
$\implies$ $PT=PB=OA = \dfrac{h}{\sqrt{3}}$ $\cdots$ (2) [From equation (1)]
From the figure $PT=h-10$
$\therefore$ From equation (2) we have
$h-10 = \dfrac{h}{\sqrt{3}}$
i.e. $h-\dfrac{h}{\sqrt{3}} = 10$
i.e. $\dfrac{\left(\sqrt{3}-1\right)h}{\sqrt{3}}=10$
i.e. $0.732h = 17.32$
$\implies$ $h = \dfrac{17.32}{0.732} = 23.66$
$\therefore$ Height of the tower is 23.66m