If the angle of elevation of the top of a chimney from a point on the horizontal plane passing through the foot of the chimney is `60^(@)` and the angle of elevation from another point on the same plane at a distance of 24 metres away from the first point is `30^(@)`. Calculate the height of the chimney. `[sqrt(3)= 1.732`(approx). ]

Explanation

Let AB be the chimney. The angle of elevation of the top A of it from the point C is `60^(@)` and from the point
` therefore " " angleACB = 30^(@) and angleADB = 60^(@)`.
Now from the right-angled triangle ABC we get,
`tan60^(@) =(AB)/(BC)` [ by definition ]
or, `sqrt(3) = (AB)/(BC)`
or, AB = `sqrt(3)BC`.................(1)
Again, from the right-angled triangle ABD we get,
`tan 30^(@) =(AB)/(BD)` [ by definition ]
or, `(1)/(sqr(3)=(AB)/(BC+CD)`
or, `sqrt(3)AB = BC + 24 [ because " "CD = 24 m ]`
or, `sqrt(3)AB - BC = 24`
or, `sqrt(3)AB - (AB)/(sqrt(3))=24` [from-(1)
or, `AB(sqrt(3)-(1)/(sqrt(3)))=24 " "or," "AB((3-1)/(sqrt(3)))=24`
or, `AB XX (2)/(sqrt(3))=24" "or, AB = (24sqrt(3))/(2)`
or, `AB = 12sqrt(3) = 12 xx 1.732` (approx.)
`therefore` Hence of the chimney = 20.784 metres (approx.)