In the given figure, triangle ABC is right-angled at B. D is the foot of the perpendicular from B to AC. Given that BC = 3 cm and AB = 4 cm. Find :

(i) tan ∠DBC

(ii) sin ∠DBA

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = 4² + 3²

⇒ AC² = 16 + 9

⇒ AC² = 25

⇒ AC = $\sqrt{25}$

⇒ AC = 5

Let CD = y and BD = x

In Δ BCD,

⇒ BC² = CD² + BD² (∵ BC is hypotenuse)

⇒ 3² = y² + x²

⇒ 9 = y² + x² …………….(1)

In Δ ABD,

⇒ AB² = AD² + BD² (∵ BC is hypotenuse)

⇒ 4² = (5 - y)² + x²

⇒ 16 = 25 + y² - 10y + x² …………….(2)

Subtracting (1) from (2), we get

⇒ 16 - 9 = (25 + y² - 10y + x²) - (y² + x²)

⇒ 7 = 25 + y² - 10y + x² - y² - x²

⇒ 7 = 25 - 10y

⇒ 7 - 25 = -10y

⇒ -10y = -18

⇒ y = $\dfrac{18}{10}$ = 1.8

AD = 5 - 1.8 = 3.2

Using equation (1),

⇒ 9 = (1.8)² + x²

⇒ x² = 9 - 3.24

⇒ x² = 5.76

⇒ x = $\sqrt{5.76}$

⇒ x = 2.4

(i) cos ∠DBC = $\dfrac{Perpendicular}{Base}$

$= \dfrac{CD}{BD}\\[1em] = \dfrac{1.8}{2.4}\\[1em] = \dfrac{3}{4}$

Hence, tan ∠DBC = $\dfrac{3}{4}$.

(ii) sin ∠DBA = $\dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{AD}{AB}\\[1em] = \dfrac{3.2}{4}\\[1em] = \dfrac{4}{5}$

Hence, sin ∠DBA = $\dfrac{4}{5}$.