A tower subtends an angle alpha at a point A | vedclass

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(B) Let the tower be $OT$ with height $H$,where $O$ is the foot of the tower. Let $A$ be a point on the ground such that $\angle OAT = \alpha$. Thus,i

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$l 	an \alpha \cot \beta$

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(B) Let the tower be $OT$ with height $H$,where $O$ is the foot of the tower. Let $A$ be a point on the ground such that $\angle OAT = \alpha$. Thus,in $	riangle AOT$,$	an \alpha = \frac{H}{AO}$,which implies $AO = H \cot \alpha$.Let $P$ be a point at height $l$ vertically above $A$,so $PA = l$. The angle of depression of the foot $O$ from $P$ is $\beta$,so $\angle APO = 90^\circ - \beta$ or $\angle POA = \beta$. In $	riangle PAO$,$	an \beta = \frac{PA}{AO} = \frac{l}{AO}$,which implies $AO = l \cot \beta$.Equating the two expressions for $AO$: $H \cot \alpha = l \cot \beta$.Therefore,$H = l \frac{\cot \beta}{\cot \alpha} = l 	an \alpha \cot \beta$.

$A$ tower subtends an angle $\alpha$ at a point $A$ in the plane of its base,and the angle of depression of the foot of the tower from a point $P$ at a height $l$ meters vertically above $A$ is $\beta$. The height of the tower is

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