Sin θ =1 2 .Find Cos θ= , Tan θ= ?

Question

Sin θ =$\frac{1}{2}$.Find Cos θ= , Tan θ= ?

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Answer

$\sin \theta=\frac{1}{2}$..(i) [Given]
In right angled $\triangle ABC$,
$\angle C=\theta \text {. }$

$\begin{aligned} \sin \theta & =\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }} \\ \therefore \quad \sin \theta & =\frac{ AB }{ AC } \quad \ldots \text { (ii) } \\ \therefore \quad \frac{ AB }{ AC } & =\frac{1}{2} \quad \ldots[\text { From (i) and (ii)] }\end{aligned}$
Let the common multiple be $k$.
$\therefore A B=1 k \text { and } BC=2 k$
Now, $A C^2=A B^2+B C^2 \ldots$ [Pythagoras theorem]
$\therefore 2 K^2=K^2+BC^2$
$\therefore 4 K^2=K^2+BC^2$
$\therefore B C^2=4 K^2-K^2=3 K^2$
$\therefore B C=\sqrt{3 k^2} \ldots . \text { [Taking square root of both sides] }$
$=\backslash \backslash \text { sqrt }\{3\{k\} \backslash$
$\therefore \quad \cos \theta=\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}=\frac{B C}{A C}=\frac{\sqrt{3} k}{2 k}=\frac{\sqrt{3}}{2}$
$\quad \tan \theta=\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}=\frac{A B}{B C}=\frac{1 k}{\sqrt{3} k}=\frac{1}{\sqrt{3}}$