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,
$AC^2 = (AB)^2 + (BC)^2$
$\Rightarrow (AB)^2 = (AC)^2 - (BC)^2$
$= 41^2 - 9^2 = 1681 - 81 = 1600$
$\Rightarrow AB = 40$
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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