The angle of elevation of the top of a tower as observed from a point on the ground is αα and on moving a metersa meters towards the tower, the angle of elevation is ββ. Prove that the height of the tower is atanαtanβtanβ−tanα.atanαtanβtanβ−tanα.
Answer:
- The situation given in the question is represented by the image given below.
Let ABAB be a tower of height hh. - In the right-angled triangle ABCABC, we have cotβ=BCBA⟹cotβ=xh⟹x=hcotβ=htanβ…(i)
- In the right-angled triangle ABD, we have cotα=BDBA⟹cotα=x+ah⟹x+a=hcotα=htanα⟹h=(x+a)tanα…(ii)
- Now, let us substitute the value of x in eq (ii). h=(htanβ+a)tanα⟹h=htanαtanβ+atanα⟹htanβ=htanα+atanαtanβ⟹h(tanβ−tanα)=atanαtanβ⟹h=atanαtanβtanβ−tanα
- Thus, the height of the tower is atanαtanβtanβ−tanα meters.