Question
Tan θ =$\frac{1}{2√2}$.Find Sin θ= , Cos θ= ?
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Answer
$\tan \theta=\frac{1}{2 \sqrt{2}}$..(i) [Given]
In right angled $\triangle A B C$,
$\angle C=\theta \text {. }$
$ \tan \theta=\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}$
$\therefore \quad \tan \theta=\frac{ AB }{ BC } \quad \ldots \text { (ii) }$
$\therefore \quad \frac{ AB }{ BC }=\frac{1}{2 \sqrt{2}} \quad \ldots[\text { From (i) and (ii)] }$
Let the common multiple be k.
\therefore $AB = 1k$ and $AC = 2√2 k$
Now, $AC^2 = AB^2 + BC^2$ …[Pythagoras theorem]
$= K^2 + (2√2 k )^2$
$= K^2 – 225^2$
$= 25K^2 + 8K^2$
$= 9K^2$
$\therefore AC =\sqrt{9 k^2} \ldots \text {.. [Taking square root of both sides] }$
$=3 K$
$\therefore \quad \sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}=\frac{ AB }{ AC }=\frac{1 k }{3 k }=\frac{1}{3}$
$\quad \cos \theta=\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}=\frac{ BC }{ AC }=\frac{2 \sqrt{2} k }{3 k }=\frac{ 2 \sqrt{2}}{3}$
In right angled $\triangle A B C$,
$\angle C=\theta \text {. }$
$ \tan \theta=\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}$
$\therefore \quad \tan \theta=\frac{ AB }{ BC } \quad \ldots \text { (ii) }$
$\therefore \quad \frac{ AB }{ BC }=\frac{1}{2 \sqrt{2}} \quad \ldots[\text { From (i) and (ii)] }$
Let the common multiple be k.
\therefore $AB = 1k$ and $AC = 2√2 k$
Now, $AC^2 = AB^2 + BC^2$ …[Pythagoras theorem]
$= K^2 + (2√2 k )^2$
$= K^2 – 225^2$
$= 25K^2 + 8K^2$
$= 9K^2$
$\therefore AC =\sqrt{9 k^2} \ldots \text {.. [Taking square root of both sides] }$
$=3 K$
$\therefore \quad \sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}=\frac{ AB }{ AC }=\frac{1 k }{3 k }=\frac{1}{3}$
$\quad \cos \theta=\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}=\frac{ BC }{ AC }=\frac{2 \sqrt{2} k }{3 k }=\frac{ 2 \sqrt{2}}{3}$
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