In the adjoining figure, AB = 4 m and ED = 3 m. If sin α = $\dfrac{3}{5}$ and cos β = $\dfrac{12}{13}$, find the length of BD.

In △ABC,

By formula,

sin α = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

Substituting values we get :

$\Rightarrow \dfrac{3}{5} = \dfrac{AB}{AC} \\[1em] \Rightarrow \dfrac{3}{5} = \dfrac{4}{AC} \\[1em] \Rightarrow AC = \dfrac{20}{3}$

In right angle triangle ABC,

By pythagoras theorem, we get :

⇒ AC² = AB² + BC²

$\Rightarrow \Big(\dfrac{20}{3}\Big)^2 = 4^2 + BC^2 \\[1em] \Rightarrow \dfrac{400}{9} = 16 + BC^2 \\[1em] \Rightarrow BC^2 = \dfrac{400}{9} - 16 \\[1em] \Rightarrow BC^2 = \dfrac{400 - 144}{9} \\[1em] \Rightarrow BC^2 = \dfrac{256}{9} \\[1em] \Rightarrow BC = \sqrt{\dfrac{256}{16}} \\[1em] \Rightarrow BC = \dfrac{16}{3} \text{ m}$

In △CDE,

By formula,

cos β = $\dfrac{\text{Base}}{\text{Hypotenuse}}$

Substituting values we get :

$\Rightarrow \dfrac{12}{13} = \dfrac{CD}{CE}$

Let CD = 12k and CE = 13k.

In right angle triangle ABC,

⇒ CE² = CD² + ED²

⇒ (13k)² = (12k)² + 3²

⇒ 169k² = 144k² + 3²

⇒ 3² = 169k² - 144k²

⇒ 9 = $25k^2$

⇒ $k = \sqrt{\dfrac{9}{25}} = \dfrac{3}{5}$.

CD = 12k = $12 \times \dfrac{3}{5} = \dfrac{36}{5}$

From figure,

$BD = BC + CD = \dfrac{16}{3} + \dfrac{36}{5} \\[1em] = \dfrac{80 + 108}{15} \\[1em] = \dfrac{188}{15} \\[1em] = 12\dfrac{8}{15} \text{ m}.$

Hence, BD = $12\dfrac{8}{15}$ m.