If cos B =5/9, then what is the value of cosec B ?

cosB = $\frac{base}{hypotenuse} = $\frac{5}{9}$

In a triangle ABC,

Base = 5

Hypotenuse = 9

Perpendicular = $\sqrt{hyp.^2 - base^2}$

                    = $\sqrt{9^2 - 5^2}$ = $\sqrt{81 - 25}$

                    = $\sqrt{56}$ = 2$\sqrt{14}$

cosecB = $\frac{hypotenuse}{perpendicular}$

           = $\frac{9}{2\sqrt{14}}$