Question
Sin θ =$\frac{11}{61}$.Find Cos θ= , Tan θ= ?
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Answer
$\sin \theta=\frac{11}{61}$
(i) [Given]
In right angled ∆ABC, ∠C = θ.
$\begin{array}{ll} & \sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }} \\ \therefore \quad & \sin \theta=\frac{ AB }{ AC } \ldots \text { (ii) } \\ \therefore \quad & \frac{ AB }{ AC }=\frac{11}{61} \quad \ldots[\text { From (i) and (ii)] }\end{array}$
Let the common multiple be k.
AB = 11k and AC = 61k
Now, $AC^2 = AB^2 + BC^2 …[Pythagoras theorem]$
$\therefore (61k)^2 = (11k)^2 + BC^2$
$\therefore 3721k^2 = 121k^2 + BC^2$
$\therefore BC^2 = 3721k^2 – 121k^2 = 3600k^2$
$BC =\sqrt{3600 k^2}$.. [Taking square root of both sides]
$=60 k$
$\begin{aligned} \therefore \quad \cos \theta & =\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}=\frac{ BC }{ AC }=\frac{60 k }{61 k }=\frac{60}{61} \\ \tan \theta & =\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}=\frac{ AB }{ BC }=\frac{11 k }{60 k }=\frac{11}{60}\end{aligned}$
(i) [Given]
In right angled ∆ABC, ∠C = θ.
$\begin{array}{ll} & \sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }} \\ \therefore \quad & \sin \theta=\frac{ AB }{ AC } \ldots \text { (ii) } \\ \therefore \quad & \frac{ AB }{ AC }=\frac{11}{61} \quad \ldots[\text { From (i) and (ii)] }\end{array}$
Let the common multiple be k.
AB = 11k and AC = 61k
Now, $AC^2 = AB^2 + BC^2 …[Pythagoras theorem]$
$\therefore (61k)^2 = (11k)^2 + BC^2$
$\therefore 3721k^2 = 121k^2 + BC^2$
$\therefore BC^2 = 3721k^2 – 121k^2 = 3600k^2$
$BC =\sqrt{3600 k^2}$.. [Taking square root of both sides]
$=60 k$
$\begin{aligned} \therefore \quad \cos \theta & =\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}=\frac{ BC }{ AC }=\frac{60 k }{61 k }=\frac{60}{61} \\ \tan \theta & =\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}=\frac{ AB }{ BC }=\frac{11 k }{60 k }=\frac{11}{60}\end{aligned}$
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