On a straight line passing through the foot of a tower, two points C and D are at distances of 4 m and 16 m from the foot respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower.
Solution
Let AB be the tower of height h m. Suppose the angle of elevation of the top of the tower from point C on the ground be θ. Then the angle of elevation of the top of the tower from point D is 90°− θ.
Here, BC = 4 m and BD = 16 m
In ΔABC
`tan theta = (AB)/(BC) = h/4` .....(1)
In ΔABD
`tan(90^@ - theta) = (AB)/(BD) = h/16`
`=> cot theta = h/16` .....(2)
Multiplying (1) and (2), we get
`tan theta xx cot theta = h/4 xx h/16 = h^2/64`
`=> 1 = h^2/64`
`=> h^2 = 64`
`=> h = 8 m`
Thus, the height of the tower is 8 m.
