In the figure (2), given below, ΔABC is right angled at B. If sin θ = $\dfrac{5}{13}$ and BC = 24 cm, find AB and AC.

Given, ΔABC is a right angled at B.

sin θ = $\dfrac{5}{13}$ and BC = 24 cm

$\Rightarrow \text{sin θ} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}}\\[1em] \Rightarrow \dfrac{5}{13} = \dfrac{AB}{AC}$

Let AB = 5k and AC = 13k.

Using pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ (13k)² = (5k)² + 24²

⇒ 169k² = 25k² + 576

⇒ 169k² - 25k² = 576

⇒ 144k² = 576

⇒ k² = $\dfrac{576}{144}$

⇒ k² = 4

⇒ k = $\sqrt{4}$

⇒ k = ± 2

Since, length cannot be negative.

⇒ k = 2

So, AB = 5k = 5 x 2 = 10 cm,

and AC = 13k = 13 x 2 = 26 cm.

Hence, AB = 10 cm and AC = 26 cm.