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The angles of elevation of the top of a tower from two points at a distance of 4m and 9m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6m.

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Let AB be the tower.
C and D be the two points with distance 4 m and 9 m from the base respectively.
In  right ΔABC,
tan x = AB/BC
⇒ tan = AB/4
⇒ AB = 4 tan x ... (i)
In  right ΔABD,
tan (90°-x) = AB/BD
⇒ cot = AB/9
⇒ AB = 9 cot  ... (ii)
Multiplying  eqn (i) and (ii)
AB2 = 9 cot × 4 tan x
⇒ AB2 = 36
⇒ AB = ± 6
Height cannot be negative. Therefore, the height of the tower is 6 m. Hence, Proved.

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