Sin θ =3 5 .Find Cos θ= , Tan θ= ?

Question

Sin θ =$\frac{3}{5}$.Find Cos θ= , Tan θ= ?

✓

Answer

$\sin \theta=\frac{3}{5}$..(i) [Given]
In right angled $\triangle ABC$,
$\angle C=\theta \text {. }$

$\sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}$
$\therefore \quad \sin \theta=\frac{ AB }{ AC } \quad \ldots \text { (ii) } $
$\therefore \quad \frac{ AB }{ AC }=\frac{3}{5} \quad \ldots[\text { From (i) and (ii)] }$
Let the common multiple be k.
$\therefore A B=3 k \text { and } A C=5 k$
Now, $AC ^2= AB ^2+ BC ^2 \ldots$ [Pythagoras theorem]
$\therefore(5) K^2=(3) K^2+BC^2$
$\therefore 25 K^2=9 K^2-225^2$
$\therefore BC^2=25 K^2-9 K^2$
$\therefore B C=\sqrt{16 k^2} \ldots \text { [Taking square root of both sides] }$
$=4 K$
$\cos \theta=\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}=\frac{ BC }{ AC }=\frac{4 k }{5 k }=\frac{4}{5}$
$\tan \theta=\frac{\text { Opposite side of } \theta}{\text { Adiacent side of } \theta}=\frac{ AB }{ BC }=\frac{3 k }{4 k }=\frac{3}{4}$