Question
If $\cos B=\frac{1}{3}$ and $\angle C=90^{\circ}$, find $\sin A$, and $B$ and $\cot A$.
✓
Answer
$\cos B=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{B C}{A B} $
$ (A B)^2=(A C)^2+(B C)^2$
$ \Rightarrow A C=\sqrt{(A B)^2-(B C)^2} $
$ \Rightarrow A C=\sqrt{3^2-1} $
$ =\sqrt{9-1} $
$=2 \sqrt{2} $
$ \sin A=\frac{B C}{A B}=\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{1}{3} $
$ \tan B=\frac{A C}{B C}=\frac{\text { Perpendicular }}{\text { Base }}=2 \sqrt{2} $
$ \cot A=\frac{1}{\tan A}=\frac{\text { BAse }}{\text { Perpendicular }}=\frac{A C}{B C}=2 \sqrt{2} .$
$ (A B)^2=(A C)^2+(B C)^2$
$ \Rightarrow A C=\sqrt{(A B)^2-(B C)^2} $
$ \Rightarrow A C=\sqrt{3^2-1} $
$ =\sqrt{9-1} $
$=2 \sqrt{2} $
$ \sin A=\frac{B C}{A B}=\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{1}{3} $
$ \tan B=\frac{A C}{B C}=\frac{\text { Perpendicular }}{\text { Base }}=2 \sqrt{2} $
$ \cot A=\frac{1}{\tan A}=\frac{\text { BAse }}{\text { Perpendicular }}=\frac{A C}{B C}=2 \sqrt{2} .$
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