If the angle, fo depression of two conseutive mile stones on a road from an aeroplane are `60^(@) and 30^(@)` respectively. Find the height of the aeroplane, (i) when the two mile stone stand on opposite side of the aeroplane, (ii) when the two stones stand on the same side of the aeroplane.

Explanation

Let O be the position of the observer. In the figure-4.34(i) the two miles are at the point A and B respectively on the opposite sides of the observer. In the figure-4.34(ii) the two miles are at the points A and B respectively on the same side of the observer.
OC is the height of the aeroplane.
Now, in figure-4.34(i), from the right-angled triangle OAC we get,
`tan 60^(@) = (OC)/(AC)`
or, `sqrt(3) = (OC)/(AC) " " or, " "OC = sqrt(3)AC " "or " " AC = (OC)/(sqrt(3))` ........ (1)
Also, from the right-angled triangle OBC we get,
`tan 30^(@) = (OC)/(BC)`
or, `(1)/(sqrt(3)) = (OC)/(BC) " " or, BC = sqrt(3)OC`..........(2)
Adding (1) and (2) we ger,
AC + BC =`(OC)/(sqrt(3)) + sqrt(3) OC`
or, AB = `OC ((1)/(sqrt(3)) + sqrt(3))`
or, 1 `= OC((1 + 3)/(sqrt(3)))" " [because` The distance between two mile stones is 1 mile ]
or, `1 = OC xx (4)/(sqrt(3)) " "or, " "OC = (sqrt(3))/(4)`
Hence the height of the aeroplane is `(sqrt(3))/(4)` mile.
Again in figure-4.34(ii), from right-angled triangle OAC we get,
`tan 60^(@) = (OC)/(AC)` [ by definition]
or, `sqrt(3)= (OC)/(AC) " "or, " "AC =(OC)/(sqrt(3))`................(1)
Also, from the right-angled triangle OBC we get,
`tan 30^(@) = (OC)/(BC)` [ by definition ]
or, `(1)/(sqrt(3)) = (OC)/(BC) " "or, " "BC = sqrt(3)OC`...................(2)
Now, adding (1) and (2) we get,
`AC + BC = (OC)/(sqrt(3)) + sqrt(3)OC`
or, `AC + AB + AC = OC((1)/(sqrt(3))+sqrt(3)) " "or, " "AB + 2AC = OC((1 + 3)/(sqrt(3)))`
or, `AB + 2 xx(OC)/(sqrt(3)) = OC xx (4)/(sqrt(3)) " "or, AB = OCxx(4)/(sqrt(3))-OCxx(2)/(sqrt(3))`
or, `1 = OCxx((4)/(sqrt(3))-(2)/(sqrt(3)))" "[because` The distance between two mile stones is 1 mile ]
or, `1-OCxx(4-2)/(sqrt(3)) or, " "1 = OCxx(2)/(sqrt(3)) " "or, OC = (sqrt(3))/(2)`
Hence the height of the aeroplane is `(sqrt(3))/(2)`mile.