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(D) Let the tower be $OP$ where $O$ is the foot of the tower on the ground and $P$ is the top. Let $h$ be the height of the tower,so $OP = h$.Since th

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$R 	an \alpha$

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(D) Let the tower be $OP$ where $O$ is the foot of the tower on the ground and $P$ is the top. Let $h$ be the height of the tower,so $OP = h$.Since the angle of elevation of the top $P$ from points $A, B, C$ is the same angle $\alpha$,we have $\angle PAO = \angle PBO = \angle PCO = \alpha$.In the right-angled triangles $\Delta POA, \Delta POB, \Delta POC$,we have $OA = OB = OC = h \cot \alpha$.This implies that $O$ is the circumcenter of $\Delta ABC$ and the distance from $O$ to the vertices $A, B, C$ is the circum-radius $R$.Therefore,$R = h \cot \alpha$.Thus,the height of the tower $h = R 	an \alpha$.

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 $\Delta ABC$,then the height of the tower is

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