Question

If $sin\theta =\frac{3}{5}$ and $\theta$ is acute angle, find
(i)$tan\theta$

Answer 1

Given:$\sin \theta =\frac{3}{5}$

$\cos \theta =\sqrt{1-{\sin }^2\theta }$

$=\sqrt{1-{\left(\frac{3}{5}\right)}^2}$

$=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{25-9}{25}}=\sqrt{\frac{16}{25}}=\pm \frac{4}{5}$

since $\theta$ is acute $\cos \theta =\frac{4}{5}$

$\therefore \tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$