As shown in the figure, AL = 20 m, BH = 80 m.
Let the height P = PQ = h metres (say)
Let ∠LBA = ϕ and ∠MAB = θ. Then,
In ΔMAB = tan θ = \(\frac{MB}{AB}\) = \(\frac{80}{AB}\) ......(i)
In ΔPQA, = tan θ = \(\frac{PQ}{QA}\) = \(\frac{h}{QA}\) ......(ii)
∴ From (i) and (ii), \(\frac{80}{AB}\) = \(\frac{h}{QA}\) ⇒ QA = \(\frac{h}{80}\). AB .....(iii)
In ΔLBA, tan ϕ, \(\frac{LA}{AB}\) = \(\frac{20}{AB}\) ......(iv)
In ΔPQB, tan ϕ = \(\frac{PQ}{BQ}\) = \(\frac{h}{BQ}\) ......(v)
∴ From (iv) and (v), \(\frac{20}{AB}\) = \(\frac{h}{BQ}\) ⇒ BQ = \(\frac{h}{20}\) AB .....(vi)
from (iii) and (vi) BQ + QA = \(\frac{h}{20}\) AB + \(\frac{h}{80}\) AB
⇒ AB = h \(\bigg(\frac{1}{20}+\frac{1}{80}\bigg)\)AB ⇒ 1 = \(\big(\frac{1+4}{80}\big)\)h ⇒ h = \(\frac{80}{5}\) = 16 m.
∴ Height of PQ from horizontal plane = 16 m.