In the figure, given below, it is given that AB is perpendicular to BD and is of length X metres. DC = 30 m, ∠ADB = 30° and ∠ACB = 45°. Find X.

From figure,

In △ABD,

$\Rightarrow \text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BD} \\[1em] \Rightarrow AB = \dfrac{BD}{\sqrt{3}} ……..(1)$

In △ABC,

$\Rightarrow \text{tan 45°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \dfrac{AB}{BC} \\[1em] \Rightarrow AB = BC ……….(2)$

From (1) and (2),

$\Rightarrow BC = \dfrac{BD}{\sqrt{3}} \\[1em] \Rightarrow \sqrt{3}BC = BD \\[1em] \Rightarrow \sqrt{3}BC = BC + CD \\[1em] \Rightarrow \sqrt{3}BC - BC = 30 \\[1em] \Rightarrow BC(\sqrt{3} - 1) = 30 \\[1em] \Rightarrow BC = \dfrac{30}{\sqrt{3} - 1} \\[1em] \Rightarrow BC = \dfrac{30}{1.732 - 1} \\[1em] \Rightarrow BC = \dfrac{30}{0.732} \\[1em] \Rightarrow BC = 40.98 \text{ meters}.$

∴ AB = X = 40.98 meters …..[From (2)]

Hence, X = 40.98 meters.