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 represent height of first building and CD represent height of second building.
BD is the width of the road.
Draw AMCD
Given Angle of elevation CAM = 60˚
Given AB = 10
BD = 12
In AMDB , D = B = 90˚
Since AMCD , M = 90˚
A = 90˚ [Angle sum property of quadrilateral]
Since each angle equal to 90 ˚ AMDB is a rectangle.
Opposite sides are equal.
AB =MD = 10
AM = BD = 12
InAMC, tan 60˚ = CM/AM
√3 = CM/12
CM = 12√3 = 20.76
CD = CM+DM
CD = 20.76 +10 = 30.76
Hence height of second building is 30.76m.