Question
If $\sin\text{A}=\frac{9}{41},$ find the value of $\cos\text{A}$ and $\tan\text{A}.$
✓
Answer
Cosider $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$
Then, $\sin\text{A}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{9}{41}$
Let $BC = 9$ and $AC = 41$
Then, by Pythagoras theoram,
$ A C^2=(A B)^2+(B C)^2$
$ \Rightarrow(A B)^2=(A C)^2-(B C)^2 $
$ =41^2-9^2=1681-81=1600$
$\Rightarrow A B=40$
Now,
$\cos\text{A}=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{40}{41}$
$\tan\text{A}=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{BC}}{\text{AB}}=\frac{9}{40}$
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