A tower subtends an angle of 30^{circ} at a p | vedclass

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(A) Let $PQ = x$ be the height of the tower and $A$ be the point on the same level as the foot of the tower $Q$. Given $\angle PAQ = 30^{\circ}$.Let $

Vedclass · Accepted Answer

$h \cot 60^{\circ}$

Vedclass · Answer

(A) Let $PQ = x$ be the height of the tower and $A$ be the point on the same level as the foot of the tower $Q$. Given $\angle PAQ = 30^{\circ}$.Let $B$ be a point $h \ m$ vertically above $A$,so $AB = h$. The angle of depression of the foot of the tower $Q$ from $B$ is $60^{\circ}$,which implies $\angle BQA = 60^{\circ}$.In the right-angled triangle $	riangle ABQ$,we have:$	an 60^{\circ} = \frac{AB}{AQ}$$\sqrt{3} = \frac{h}{AQ}$$AQ = \frac{h}{\sqrt{3}} = h \cot 60^{\circ}$.Thus,the horizontal distance of the tower from the point $A$ is $h \cot 60^{\circ}$.

$A$ tower subtends an angle of $30^{\circ}$ at a point on the same level as the foot of the tower. At a second point,$h \ m$ above the first,the angle of depression of the foot of the tower is $60^{\circ}$. The horizontal distance of the tower from the point is

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