ABCD is a rhombus with diagonals BD and AC equal to 12 cm and 16 cm respectively. The value of cosec x is :

1. $1\dfrac{2}{3}$
2. $3\dfrac{1}{3}$
3. $\dfrac{1}{2}$
4. $1\dfrac{1}{3}$

The diagonals of a rhombus bisect each other at right angles.

∴ ∠AOB = 90°

In Δ AOB,

⇒ AB² = BO² + OA² (∵ AB is hypotenuse)

⇒ AB² = 6² + 8²

⇒ AB² = 36 + 64

⇒ AB² = 100

⇒ AB = $\sqrt{100}$

⇒ AB = 10 cm

cosec $x = \dfrac{Hypotenuse}{Perpendicular}$

$= \dfrac{AB}{OB}\\[1em] = \dfrac{10}{6}\\[1em] = \dfrac{5}{3}\\[1em] = 1\dfrac{2}{3}$

Hence, option 1 is the correct option.