The angle of elevation of the highest point P of a vertical tower at point A on the horizontal ground is `45^(@)`. The height of tower is 'h'/ The angle of a elevation of the tower becomes `60^(@)` from B on moving a distance 'd' at `30^(@)` angle from the horizontal. Prove that: `d=h(sqrt(3)-1)`

Explanation

Here AB=d and PQ=h
`anglePAQ=45^(@), angleBAQ=30^(@)` and `anglePBH=60^(@)`

`therefore angleBAP=15^(@)`
Now `angleAPQ=45^(@)`
and `angleBPQ=30^(@)`
`therefore angleAPB=15^(@)`
`rArr angleABP=180^(2)-15^(2)=150^(@)`
Now PQ=AQ=h
`AP^(2)=AQ^(2)+PQ^(2)`
`=h^(2)+h^(2)=2h^(2)`
`rArr AP=hsqrt(2)`
In `DeltaABP`
`(AB)/(sin15^(@))=(AP)/(sin 150^(@))`
`rArr d/(sin15^(@))=(hsqrt(2))/(sin30^(@))`
`rArr d=(hsqrt(2).sin15^(@))/(sin30^(@))`
`rArr d=((hsqrt(2)(sqrt(3)-1))/(2sqrt(2)))/(1/2)`
`=h(sqrt(3)-1)` Hence Proved.