Two lamp posts AB and CD stand on a levelled ground and are of equal heights. A boy measured the elevation of the top lamp post AB from the midpoint M of the line segment joining the foot of the lamp posts as 30 degrees. After walking 15 metres towards lamp post CD he measured the elevation of its tops 60 degrees. Find the distance between the lamp posts

Solution :

                    Let BM= x metres. 

Since M is the midpoint of BD.

                    BD = 2x metres  

Given that 

                    MN =15 metres. 

                    ND = (x-15) metres. 

In triangle ABM

                    Tan 30 degrees = AB/x

                    1/√3 = AB/x

                    AB  = x/√3 ---------------         (1)

In triangle CDN

                    Tan 60 degrees = CD/ND

                     √3 = CD/(x-15)

                    CD = √3 (x-15) -----------        (2)

Given That

                    AB = CD.        

Hence  

                    x/√3 = √3 (x - 15 )

                    x = 3 (x - 15)

                    x = 3x - 45

                    45 = 3x -1x

                    45 = 2x

                    45/2 = x

                    22.5 = x

Therefore 

The distance between the lamp posts is BD

                    BD = 2x

                    BD = 2 * 22.5

                    BD = 45 metres

The angle of elevation of the top of a building CD from the foot of the tower AB is 30 degrees and the angle of elevation of the top of the tower from the foot of the building is 60 degrees .If the tower is 50 m high. Find the height of the building. ( (Take √3 = 1.73 )

In triangle ABC

Cot 60 degrees  = BC/AB = Adjacent Side/opposite side. 

1/√3 = BC/50 .

50 = BC (√3 ) .

50/√3 = BC .

In triangle BCD

Tan 30 degrees  = DC/BC = Opposite side / Adjacent side. 

1/√3 = DC/ (50/√3 )

1/√3 = DC (√3/50 )

 50 = DC ( 3 )

DC = 50/3

DC = 16.7 m

The height of the building is 16.7 metres.