In the given figure; BC = 15 cm and $\text{sin B} = \dfrac{4}{5}$.

(i) Calculate the measures of AB and AC.

(ii) Now, if tan ∠ADC = 1; calculate the measures of CD and AD.

(i) Given:

$\text{sin B} = \dfrac{4}{5}\\[1em] \text{sin B} = \dfrac{Perpendicular}{Hypotenuse} = \dfrac{4}{5}$

∴ If length of AC = 4x cm, length of AB = 5x cm.

In Δ ABC,

⇒ AB²= BC² + AC² (∵ AB is hypotenuse)

⇒ (5x)² = BC² + (4x)²

⇒ 25x² = BC² + 16x²

⇒ BC² = 25x² - 16x²

⇒ BC² = 9x²

⇒ BC = $\sqrt{9\text{x}^2}$

⇒ BC = 3x

It is given that BC = 15 cm

3x = 15

x = $\dfrac{15}{3}$

x = 5 cm

AB = 5x = 5 x 5 cm = 25 cm

AC = 4x = 4 x 5 cm = 20 cm

Hence, AB = 25 cm and AC = 20 cm.

(ii) tan ∠ADC = 1

$\text{tan ∠ADC} = \dfrac{Perpendicular}{Base} = 1$

∴ If length of AC = x unit, length of CD = x unit.

From (i), we know AC = 20 cm

∴ x = 20 cm

So, AC = CD = 20 cm

In Δ ACD,

⇒ AD² = AC² + CD² (∵ AD is hypotenuse)

⇒ AD² = 20² + 20²

⇒ AD² = 400 + 400

⇒ AD² = 800

⇒ AD = $\sqrt{800}$

⇒ AD = $20\sqrt{2}$

Hence, CD = 20 cm and AD = 20 $\sqrt{2}$ cm.