Two persons on the same side of a tall building notice the angle of elevation of the top of the building to be `30^(@)` and `60^(@)` respectively. If the height of the building is `72m`, find the distance between the two persons to the nearest metre. `(sqrt(3)=1.73`)

Explanation

In the figure,
AB is the building of height `72 m`.
C and D are the positions of two persons `B-C-D`
Let `BC=xm` and `CD=ym`
The angle of elevation from C is `60^(@)`
In right angled `DeltaABC, tan /_ACB=tan 60^(@)=(AB)/(BC)=72/x`
`:.sqrt(3)=72/x`
`:.x=72/(sqrt(3))=72/(sqrt(3))xx(sqrt(3))/(sqrt(3))=(72sqrt(3))/3 :. x=24 sqrt(3)m`.......1

The anglee of elevation from D is `30^(@)`.
In right angled `DeltaABD, tan /_ADB=tan30^(@)=(AB)/(BD)=72/(x+y)`
`:.1/(sqrt(3))=72/(24 sqrt(3+y))`[From 1]
`:24 sqrt(3)+y=72sqrt(3)`
`:.y=72sqrt(3)-24sqrt(3)=48sqrt(3)`
`:.y=48xx1.73 :. y=83.04m~~83m`
The distance between the two persons is `83m`.