The angle of elevation of the top of a tower from a point A on the ground is 30°. Moving a distance of 20metres towards the foot of the tower to a point B the angle of elevation increases to 60°. Find the height of the tower & the distance of the tower from the point A.
Solution
Let h be the height of the tower and the angle of elevation of the top of the tower from a point A on the ground is 30° and on moving with distance 20 m towards the foot of tower on the point B is 60°.
Let AB = 20 and BC = x
Now we have to find the height of tower and distance of tower from point A.
So we use trigonometrical ratios.
In ΔDBC
`=> tan D = (CD)/(BC)`
`=> tan 60^@ = (CD)/(BC)
`=> sqrt3 = h/x`
`=> x = h/sqrt3`
Again in Δ DAC
`=> tan A = (CD)/(BC + BA)`
`=> tan 30^@ = h/(x + 20)`
`=> 1/sqrt3 = h/(x + 20)`
`=> x = sqrt3h - 20`
`=> h/sqrt3 + 20 = sqrt3h`
`=> h/sqrt3 - sqrt3h = -20`
`=> h - 3h = -20sqrt3`
`=> -2h = -20sqrt3`
`=> h = 10sqrt3`
`=> h = 17.32`
`=> x = (10sqrt3)/sqrt3`
`=> x = 10`
So distance
`=> AC = x + 20`
=> AC = 30
Hence the required height is 17.32 m and distance is 30 m
