Question
If $8 \tan A = 15,$ find $\sin A - \cos A.$
✓
Answer
$8 \tan A =15 $
$\Rightarrow \tan A =\frac{15}{8}=\frac{\text { Perpendicular }}{\text { Base }} $
Hypotenuse
$=\sqrt{(\text { Perpendicular })^2+(\text { Base })^2} $
$=\sqrt{(15)^2+(8) 2} $
$=\sqrt{225+64} $
$=\sqrt{289} $
$=17 $
$\sin A -\cos A =\frac{\text { Perpendicular }}{\text { Hypotenuse }}-\frac{\text { Base }}{\text { Hypotenuse }} $
$=\frac{15}{17}-\frac{8}{17} $
$=\frac{15-8}{17} $
$\sin A -\cos A =\frac{7}{17} .$
$\Rightarrow \tan A =\frac{15}{8}=\frac{\text { Perpendicular }}{\text { Base }} $
Hypotenuse
$=\sqrt{(\text { Perpendicular })^2+(\text { Base })^2} $
$=\sqrt{(15)^2+(8) 2} $
$=\sqrt{225+64} $
$=\sqrt{289} $
$=17 $
$\sin A -\cos A =\frac{\text { Perpendicular }}{\text { Hypotenuse }}-\frac{\text { Base }}{\text { Hypotenuse }} $
$=\frac{15}{17}-\frac{8}{17} $
$=\frac{15-8}{17} $
$\sin A -\cos A =\frac{7}{17} .$
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