In the adjoining figure, cosec x = $\dfrac{13}{5}$, AB = 26 cm and sin y = $\dfrac{8}{17}$. Find BC.

By formula,

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

cosec θ = $\dfrac{\text{Hypotenuse}}{\text{Perpendicular}}$

In ΔABD,

$\Rightarrow \text{cosec x} = \dfrac{AB}{BD}\\[1em] \Rightarrow \dfrac{13}{5} = \dfrac{26}{BD}\\[1em] \Rightarrow BD = \dfrac{26 \times 5}{13}\\[1em] \Rightarrow BD = 2 \times 5\\[1em] \Rightarrow BD = 10 \text{ cm}.$

Since, ΔABD is a right angled triangle. Using pythagoras theorem,

⇒ AB² = BD² + AD²

⇒ 26² = 10² + AD²

⇒ 676 = 100 + AD²

⇒ AD² = 676 - 100

⇒ AD² = 576

⇒ AD = $\sqrt{576}$

⇒ AD = ± 24 cm

As length of side of a triangle cannot be negative. So, AD = 24 cm.

In ΔADC,

$\Rightarrow \text{sin y} = \dfrac{AD}{AC}\\[1em] \Rightarrow \dfrac{8}{17} = \dfrac{24}{AC}\\[1em] \Rightarrow AC = \dfrac{24 \times 17}{8}\\[1em] \Rightarrow AC = 3 \times 17\\[1em] \Rightarrow AC = 51 \text{ cm}.$

Since, ΔADC is a right angled triangle. Using pythagoras theorem,

⇒ AC² = AD² + DC²

⇒ 51² = 24² + DC²

⇒ 2601 = 576 + DC²

⇒ DC² = 2601 - 576

⇒ DC² = 2025

⇒ DC = $\sqrt{2025}$

⇒ DC = ± 45

As length of side of a triangle cannot be negative. So, DC = 45 cm.

From figure,

BC = BD + DC = 10 + 45 = 55 cm.

Hence, the length of BC = 55 cm.