Question

In $\Delta ABC$ $\angle C=90$. $AC=15cm$, $BC=8cm$, find $\sec A$.

• A. $\frac{5}{17}$
• B. $\frac{17}{15}$
• C. $\frac{15}{17}$
• D. $\frac{17}{5}$

Answer 1

The correct option is B $\frac{17}{15}$

REF.Image.

In $△ABC\angle C={90}^{\circ },AC=15cm,BC=8cm,$ find secA.

$\to$ for given $△ABC,\angle C={90}^{\circ }$

using pythagorus theroem

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

$\therefore A{B}^2=(15{)}^2+(8{)}^2$

$\therefore A{B}^2=225+64$

$\therefore A{B}^2=289$

$\therefore AB=\sqrt{289}$

$\therefore AB=17cm$

$\to secA=\frac{AB}{AC}$

$\therefore secA=\frac{17}{15}$