Question
If $15\cot\text{A}=8,$ find the value of $\sin\text{A}$ and $\sec\text{A}.$
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Answer
$15\cot\text{A}=8\Rightarrow\cot\text{A}=\frac{8}{15}$ Cosider $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$ Then, $\cot\text{A}=\frac{\text{Base}}{\text{Perpendicular}}=\frac{\text{AB}}{\text{BC}}=\frac{8}{15}$ Let AB = 8 and BC = 15 Then, by Pythagoras theoram, $\text{AC}^2 = (\text{AB})^2 + (\text{BC})^2 $ $\Rightarrow(\text{AB})^2 = (\text{AC})^2 - (\text{BC})^2 $ $= 82 + 152 = 64 + 225 = 289$ $\Rightarrow \text{AC} = 17$ Now, $\sin\text{A}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{15}{17}$ $\sin\text{A}=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{\text{AC}}{\text{AB}}=\frac{17}{8}$Need a full question paper?
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