Question
If $\sin \theta=\frac{4}{5}$, then find $\cos \theta$.
✓
Answer
$\sin \theta=\frac{4}{5} \ldots$. (i) [Given]
In right angled $\triangle ABC$,
$\angle C=\theta \text {. }$
$\sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}$
$\therefore \quad \sin \theta=\frac{ AB }{ AC } \ldots \text { (ii) }$
$\therefore \quad \frac{ AB }{ AC }=\frac{4}{5}$
Let the common multiple be k.
$\therefore AB = 4k$ and $AC = 5k$
Now, $AC^2 = AB^2 + BC^2$ … [Pythagoras theorem]
$\therefore (5 k)^2 = (4k)^2 + BC^2$
$\therefore 25k^2 = 16k^2 + BC^2$
$\therefore BC^2 = 25k^2 – 16k^2 = 9k^2$
$\therefore BC =\sqrt{9 k^2} . \text {.[Taking square root of both sides] }$
$= 3 k$
$ \therefore \quad \cos \theta=\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}=\frac{ BC }{ AC }=\frac{3 k }{5 k }=\frac{3}{5}$
In right angled $\triangle ABC$,
$\angle C=\theta \text {. }$
$\sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}$
$\therefore \quad \sin \theta=\frac{ AB }{ AC } \ldots \text { (ii) }$
$\therefore \quad \frac{ AB }{ AC }=\frac{4}{5}$
Let the common multiple be k.
$\therefore AB = 4k$ and $AC = 5k$
Now, $AC^2 = AB^2 + BC^2$ … [Pythagoras theorem]
$\therefore (5 k)^2 = (4k)^2 + BC^2$
$\therefore 25k^2 = 16k^2 + BC^2$
$\therefore BC^2 = 25k^2 – 16k^2 = 9k^2$
$\therefore BC =\sqrt{9 k^2} . \text {.[Taking square root of both sides] }$
$= 3 k$
$ \therefore \quad \cos \theta=\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}=\frac{ BC }{ AC }=\frac{3 k }{5 k }=\frac{3}{5}$
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