Tan θ =8 15 .Find Sin θ= , Cos θ= ?

Question

Tan θ =$\frac{8}{15}$.Find Sin θ= , Cos θ= ?

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Answer

$\tan \theta=\frac{8}{15}$..(i) [Given]
In right angled $\triangle ABC$,
$\angle C=\theta \text {. }$

$\begin{aligned} & \tan \theta=\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta} \\ \therefore \quad \tan \theta & =\frac{ AB }{ BC } \quad \ldots \text { (ii) } \\ \therefore \quad & \frac{ AB }{ BC }=\frac{8}{15} \quad \ldots[\text { From (i) and (ii)] }\end{aligned}$
Let the common multiple be k.
Let the common multiple be $k$.
$\therefore A B=8 k \text { and } B C=15 k$
Now, $A C^2=A B^2+B C^2 \ldots[$ Pythagoras theorem $]$
$=(8) K^2+(15 K)^2$
$=64 K^2-225^2$
$=289 K^2$
$\begin{aligned}
\therefore & A C=\sqrt{289 k^2} \ldots \text {.[Taking square root of both sides] } \\
= & 17 K \\
& \therefore \quad \sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}=\frac{ AB }{ AC }=\frac{8 k }{17 k }=\frac{8}{17} \\
& \cos \theta=\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}=\frac{ BC }{ AC }=\frac{15 k }{17 k }=\frac{15}{17}
\end{aligned}$