If $\text{tan x°} = \dfrac{5}{12}$, $\text{tan y°} = \dfrac{3}{4}$ and AB = 48 m; find the length of CD.

Given:

$\text{tan x°} = \dfrac{5}{12}$, $\text{tan y°} = \dfrac{3}{4}$ and AB = 48 m

Let BC be x m.

AC = AB + BC = (48 + x) m.

In Δ ADC,

$\text{tan x°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ \dfrac{5}{12} = \dfrac{DC}{AC}\\[1em] ⇒ \dfrac{5}{12} = \dfrac{DC}{48 + x}\\[1em] ⇒ \text{12 CD = 240 + 5x} …………….(1)$

And, in Δ BDC,

$\text{tan y°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ \dfrac{3}{4} = \dfrac{DC}{BC}\\[1em] ⇒ \dfrac{3}{4} = \dfrac{DC}{x}\\[1em] ⇒ 4 CD = 3x \\[1em] ⇒ x = \dfrac{4 CD}{3}…………….(2)$

From equation (1),

240 + 5 $\times \dfrac{4CD}{3}$ = 12 CD

⇒ 240 + $\dfrac{20CD}{3}$ = 12 CD

⇒ 720 + 20 CD = 36 CD

⇒ 720 = 36 CD - 20 CD

⇒ 16 CD = 720

⇒ CD = $\dfrac{720}{16}$

⇒ CD = 45

Hence, CD = 45 m.