Correct option is: D) 120(√3 – 1) m
Let \(\angle\)CAB =\(45^\circ\) & \(\angle\)DAB = \(60^\circ\)
Also let BC be the original height of the tower and CD be the length to be raised to make angle of elevation \(60^\circ\) at point A.
In right \(\triangle\)ABC,
tan A = \(\frac {BC}{AB}\)
= BC = AB tan A = 120 tan \(45^\circ\) (\(\because\) \(\angle\)CAB =\(45^\circ\))
= BC = 120 m ....(1) (\(\because\) tan \(45^\circ\) = 1)
In right \(\triangle\)ABC,
tan A = \(\frac {BD}{AB}\)
= BD = AB tan A = AB tan \(60^\circ\) (\(\because\) \(\angle\)DAB \(60^\circ\))
= BD = 120\(\sqrt3\) ...(2) (\(\because\) AB = 120 m)
Now, CD = BD-BC
= 120\(\sqrt3\) -120 (From (1) & (2)
= 120 (\(\sqrt3\)-1)m
Hence, tower to be raised 120 (\(\sqrt3\)-1)m to make angle of elevation \(60^\circ\) at the same point.