Question

In a $\Delta ABC,\angle B={90}^o,AB=12cm$ and $BC=5cm.$ Find the value of $cosecC$ and $secA$.

• A. $\frac{12}{13},\frac{13}{12}$
• B. $\frac{5}{13},\frac{5}{12}$
• C. $\frac{13}{12},\frac{13}{12}$
• D. $\frac{5}{13},\frac{13}{12}$

Answer 1

The correct option is C $\frac{13}{12},\frac{13}{12}$

Triangle ABC is a right angled triangle.

By applying Pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2$

$AC=\sqrt{(A{B}^2+B{C}^2)}$

$AC=\sqrt{({12}^2+5^2)}$

$AC=\sqrt{144+25}=\sqrt{169}=13$

$cosecC=\frac{Hypotenuse}{Oppositeside}=\frac{AC}{AB}=\frac{13}{12}$
$secA=\frac{Hypotenuse}{Adjacentside}=\frac{AC}{AB}=\frac{13}{12}$