The angle of elevation of the top of the towe | vedclass

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(D) Let $P$ be the top of the tower and $O$ be the foot of the tower on the ground. Since the tower makes equal angles $\alpha$ at the vertices $A, B,

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$R 	an \alpha$

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(D) Let $P$ be the top of the tower and $O$ be the foot of the tower on the ground. Since the tower makes equal angles $\alpha$ at the vertices $A, B, C$ of the triangle,the foot of the tower $O$ is the circumcentre of the triangle $ABC$.In $\Delta OAP$,$\angle OAP = 90^\circ$ is not correct,rather $\angle AOP = 90^\circ$ because $OP$ is the height of the tower perpendicular to the ground.Thus,in the right-angled triangle $\Delta OAP$,we have:$	an \alpha = \frac{OP}{OA}$Since $O$ is the circumcentre,$OA = OB = OC = R$ (where $R$ is the circum-radius).Therefore,$	an \alpha = \frac{OP}{R}$$OP = R 	an \alpha$Hence,the height of the tower is $R 	an \alpha$.

The angle of elevation of the top of the tower observed from each of the three points $A, B, C$ on the ground,forming a triangle,is the same angle $\alpha$. If $R$ is the circum-radius of the triangle $ABC$,then the height of the tower is

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