The angle of elevation of to be top point `P` of the vertical tower PQ of height `h` from poin A is `45^0` and from a point B, the angle of elevation is `60^0,` where B is point at a distance `d` from the point A measured along the line `A B` which makes an angle `30^0` with AQ. Prove that `d=(sqrt(3)-1)hdot`

Explanation

Given, `PQ=h, PAQ=45^@, BAQ=30^@, AB=d. PBH=60^@`

From `Delta APQ`, we have `AP^2=h^2+h^2=2h^2`

`AQ=PQ=h`

`AP=sqrt(2)`

Applying `sin` formula in `Delta ABP`, we get,

`frac{AB}{sin APB}=frac{AP}{sin APB}`