Question
If $\sin\text{A}=\frac{9}{41},$ computer cos A and tan A.
✓
Answer
$\sin\text{A}=\frac{9}{41},$ $\sin\text{A}=\frac{\text{opposite side}}{\text{adjacent side}}=\frac{9}{41}$
Consider right angled triangle ABC,
Let x be the adjacent side By applying Pythagoras $\text{AC}^2=\text{AB}^2+\text{BC}^2$ $(41)^2=(\text{AB})^2+9^2$ $1681-81=\text{AB}^2$ $\text{AB}=40$ $\cos\text{A}=\frac{\text{adjacent side}}{\text{hypotenuse}}=\frac{40}{41}$ $\tan\text{A}=\frac{\text{opposite side}}{\text{Hypotenuse side}}=\frac{9}{40}$
Consider right angled triangle ABC,
Let x be the adjacent side By applying Pythagoras $\text{AC}^2=\text{AB}^2+\text{BC}^2$ $(41)^2=(\text{AB})^2+9^2$ $1681-81=\text{AB}^2$ $\text{AB}=40$ $\cos\text{A}=\frac{\text{adjacent side}}{\text{hypotenuse}}=\frac{40}{41}$ $\tan\text{A}=\frac{\text{opposite side}}{\text{Hypotenuse side}}=\frac{9}{40}$
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