The angle of elevation of the top of a tower at a point on the line through the foot of the tower is `45^@`. After walking a distance towards the foot of the tower along the same horizontal line elevation of the top of the tower changes to `60^@`. Find the height of tower.

Explanation

Let the height of the tower be `h` m.
Let QA= x m
`tan45^(@)rArr1=(h)/(20+x)rArr20+x=h`
`rArrx=h-20" "(1)`
From `Delta QAB`,
`tan60^(@)=(AB)/(QA)`
`(sqrt3=(h)/(x)rArrh=sqrt3" x " rArrh=sqrt3(h-20)" "(using Eq. (1))`
`rArr(sqrt3-1)h=20sqrt3rArrh=(20sqrt3)/(sqrt3-1)=(20sqrt3(sqrt3+1))/((sqrt3-1)(sqrt3+1))=(20(3+sqrt3))/(3-1)=10(3+sqrt3)`
Hence, the height of the tower is `10(3+sqrt3)` m.