Question

If $sinA=\frac{3}{4}$, then, cos A = x, and tan A = y.
Choose the correct options

• A. $x=\frac{\sqrt{7}}{4}$
  
  
• B. $y=\frac{\sqrt{7}}{4}$
  
• C. $y=\frac{3}{\sqrt{7}}$
  
• D. $x=\frac{3}{\sqrt{7}}$

Answer 1

Answer: (A), (C)

The correct options are
A $x=\frac{\sqrt{7}}{4}$


C $y=\frac{3}{\sqrt{7}}$


Given
$sinA=\frac{3}{4}$

Let us consider $△ABC$ with $\angle B={90}^{\circ }$

$sinA=\frac{BC}{AC}=\frac{3}{4}$

Let $BC=3k$. $AC=4k$

From Pythagoras theorem,

${AB}^2+{BC}^2={AC}^2$

$\Rightarrow {AB}^2+{\left(3k\right)}^2={\left(4k\right)}^2$

$\Rightarrow {AB}^2=16k^2-9k^2=7k^2$

$\Rightarrow AB=\sqrt{7}k$

$\therefore cosA=\frac{AB}{AC}=\frac{\sqrt{7}k}{4k}=\frac{\sqrt{7}}{4}$

$and,tanA=\frac{BC}{AB}=\frac{3k}{\sqrt{7}k}=\frac{3}{\sqrt{7}}$

Thus, $x=\frac{\sqrt{7}}{4},andy=\frac{3}{\sqrt{7}}$