Question

If $\tan \theta =\frac{2}{3}(0^o<\theta <{90}^o)$, then find the value of $\sin \theta$

• A. $\frac{2}{\sqrt{13}}$
• B. $\frac{3}{\sqrt{12}}$
• C. $1$
• D. $0$

Answer 1

Answer: (A)

$\frac{2}{\sqrt{13}}$

As $0^o<\theta <{90}^o$ , therefore the value of all the T-ratios have positive values
$\therefore {\sec }^2\theta =1+{\tan }^2\theta$
$\therefore {\sec }^2\theta =1+{\left(\frac{2}{3}\right)}^2$ ......$[\because \tan \theta =\frac{2}{3}]$
$=1+\frac{4}{9}=\frac{13}{9}$
$\therefore \sec \theta =\pm \sqrt{\frac{13}{9}}=\frac{\sqrt{13}}{3}$
Taking only the positive sign
$\therefore \cos \theta =\frac{3}{\sqrt{13}}$
$\therefore {\sin }^2\theta +{\cos }^2\theta =1$
$\therefore {\sin }^2+{\left(\frac{3}{\sqrt{13}}\right)}^2=1$ or ${\sin }^2\theta +\frac{9}{13}=1$
$\therefore {\sin }^2\theta =1-\frac{9}{13}=\frac{4}{13}$
$\therefore \sin \theta =\pm \sqrt{\frac{4}{13}}=\frac{2}{\sqrt{13}}$
Taking only the positive sign.