The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
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 x = AB/4
⇒ AB = 4 tan x ... (i)
also,
In right ΔABD,
tan (90°-x) = AB/BD
⇒ cot x = AB/9
⇒ AB = 9 cot x ... (ii)
Multiplying eqn (i) and (ii)
AB2 = 9 cot x × 4 tan x
⇒ AB2 = 36
⇒ AB = ± 6
Height cannot be negative. Therefore, the height of the tower is 6 m. Hence, Proved.
