Question
Cos θ =$\frac{1}{√3}$.Find Sin θ= , Tan θ= ?
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Answer
$\cos \theta=\frac{1}{\sqrt{3}}$..(i) [Given]
In right angled $\triangle A B C$, $\angle C=\theta$.
$\begin{aligned} \cos \theta & =\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }} \\ \therefore \quad \cos \theta & =\frac{ BC }{ AC } \quad \ldots \text { (ii) } \\ \therefore \quad \frac{ BC }{ AC } & =\frac{1}{\sqrt{3}} \quad \ldots[\text { From (i) and (ii)] }\end{aligned}$
Let the common multiple be $k$.
$\therefore A B=1 k \text { and } B C=\sqrt{ } 3 k$
Now, $A C^2=A B^2+B C^2 \ldots[$ Pythagoras theorem $]$
$\therefore(\sqrt{ } 3 K)^2=A B^2+K^2$
$\therefore 3 K^2=3 K^2-K^2=2 K^2$
$\therefore A B=\sqrt{2 k^2} \text {.. [Taking square root of both sides] }$
$A B=\sqrt{ } 2 K$
$\therefore \quad \sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}=\frac{A B}{A C}=\frac{\sqrt{2} k}{\sqrt{3} k}=\frac{\sqrt{2}}{\sqrt{3}}$
$\tan \theta=\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}=\frac{A B}{B C}=\frac{\sqrt{2} k }{1 k }=\sqrt{2}$
In right angled $\triangle A B C$, $\angle C=\theta$.
$\begin{aligned} \cos \theta & =\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }} \\ \therefore \quad \cos \theta & =\frac{ BC }{ AC } \quad \ldots \text { (ii) } \\ \therefore \quad \frac{ BC }{ AC } & =\frac{1}{\sqrt{3}} \quad \ldots[\text { From (i) and (ii)] }\end{aligned}$
Let the common multiple be $k$.
$\therefore A B=1 k \text { and } B C=\sqrt{ } 3 k$
Now, $A C^2=A B^2+B C^2 \ldots[$ Pythagoras theorem $]$
$\therefore(\sqrt{ } 3 K)^2=A B^2+K^2$
$\therefore 3 K^2=3 K^2-K^2=2 K^2$
$\therefore A B=\sqrt{2 k^2} \text {.. [Taking square root of both sides] }$
$A B=\sqrt{ } 2 K$
$\therefore \quad \sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}=\frac{A B}{A C}=\frac{\sqrt{2} k}{\sqrt{3} k}=\frac{\sqrt{2}}{\sqrt{3}}$
$\tan \theta=\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}=\frac{A B}{B C}=\frac{\sqrt{2} k }{1 k }=\sqrt{2}$
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