Question

In a right triangle ABC, $\angle$ B = ${90}^{\circ }$ and sec A = $\frac{5}{3}$. Find sin A.

• A. $\frac{5}{3}$
• B. $\frac{5}{4}$
• C. $\frac{3}{5}$
• D. $\frac{4}{5}$

Answer 1

The correct option is D $\frac{4}{5}$

Consider triangle ABC as shown below:

Given: $secA=\frac{5}{3}$ --- (1)

$\because secA=\frac{Hypotenuse}{Adjacentside}=\frac{AC}{AB}$ ---(2)

On comparing (1) and (2), we get

if $AC=5k$, then $AB=3k$ where k is positive number.

Using Pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2$
$\Rightarrow (5k{)}^2=(3k{)}^2+B{C}^2$
$\Rightarrow 25k^2=9k^2+B{C}^2$
$\Rightarrow BC=\sqrt{16k^2}$
$\therefore BC=4k$

Now,
$sinA=\frac{Oppositeside}{Hypotenuse}$

= $\frac{BC}{AC}$ = $\frac{4k}{5k}=\frac{4}{5}$

$\therefore sinA=\frac{4}{5}$