Two buildings are facing each other on a road of width 12 metre. From the top of the first building, which is 10 metre high, the angle of elevation of the top of the second is found to be 60°. What is the height of the second building ?
Let AB and CD be two building, with
AB = 10 m
And angle of elevation from top of AB to top of CD = ∠CAP = 60°
Width of road = BD = 12 m
Clearly, ABDP is a rectangle
AB = PD = 10 m
BD = AP = 12 m
And APC is a right-angled triangle, In ∆APC
`tan theta = "Perpendicular"/"Base" = "CP"/"AP"`
`=> tan 60° = "CP"/12`
`=> sqrt3 = "CP"/12`
⇒ CP = 12√3 m
CD = CP + PD = (12√3 + 10) m
Hence, height of other building is (10 + 12√3 m).