Question
If $\sin A=\frac{3}{5}$, find $\cos A$ and $\tan A$.
✓
Answer
$\sin A=\frac{3}{5}=\frac{\text { Perpendicular }}{\text { Hypotenuse }} $
By Pythagoras theorem, we have
$\Rightarrow($Hypotenuse$)^2=($Perpendicular$)^2+($Base$)^2 $
$\Rightarrow($Base$) ^2=($Hypotenuse$a)^2-($Perpendicular$)^2 $
$\Rightarrow(\text { Base })=\sqrt{(\text { Hypotenuse })^2-(\text { Perpendicular })^2} $
$\Rightarrow ($Base$)$
$=\sqrt{5^2-3^2} $
$=\sqrt{25-9} $
$=\sqrt{16} $
$=4 $
$\cos A =\frac{\text { Base }}{\text { Hypotenuse }}=\frac{4}{5} $
$\tan A=\frac{\text { Perpendicular }}{\text { Base }}=\frac{3}{4} \text {. }$
By Pythagoras theorem, we have
$\Rightarrow($Hypotenuse$)^2=($Perpendicular$)^2+($Base$)^2 $
$\Rightarrow($Base$) ^2=($Hypotenuse$a)^2-($Perpendicular$)^2 $
$\Rightarrow(\text { Base })=\sqrt{(\text { Hypotenuse })^2-(\text { Perpendicular })^2} $
$\Rightarrow ($Base$)$
$=\sqrt{5^2-3^2} $
$=\sqrt{25-9} $
$=\sqrt{16} $
$=4 $
$\cos A =\frac{\text { Base }}{\text { Hypotenuse }}=\frac{4}{5} $
$\tan A=\frac{\text { Perpendicular }}{\text { Base }}=\frac{3}{4} \text {. }$
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