(b) 30 ft
Let BD be the height of the flag. Given, DC = 20 ft. and length of ladder AC = 20 ft.
∠DAB = 60°.
Let BC = h metres
In rt. Δ ABD, tan 60° = \(\frac{BD}{AB}\)
⇒ √3 = \(\frac{h+20}{AB}\)
⇒ AB = \(\frac{h+20}{\sqrt3}\) = \(\frac{(h+20)\sqrt3}{3}\) .....(i)
In rt. Δ ABC, AC2 = AB2 + BC2
⇒ 202 = \(\frac{3(h+20)^2}{9}+h^2\) (From (i))
⇒ 400 = \(\frac{(h+20)^2+3h^2}{3}\) ⇒ 1200 = h2 + 40h + 400 + 3h2
⇒ 4h2 + 40h – 800 = 0 ⇒ h2 + 10h – 200 = 0
(h + 20) (h – 10) = 0 ⇒ h = 10 (∵ h ≠ – 20)
∴ Height of the flag = BC + CD = 10 + 20 = 30 ft.