Question
Given 15 cot A = 8, find sin A and sec A.
✓
Answer
We have, $15\cot\text{A}=8$ $\cot\text{A}=\frac{8}{15}=\frac{\text{Base}}{\text{Perpendicular}}$
In $\triangle\text{ABC},$ $\text{AC}^2=\text{AB}^2+\text{BC}^2$ $\Rightarrow\text{AC}^2=(8)^2+(15)^2$ $\Rightarrow\text{AC}^2=64+225$ $\Rightarrow\text{AC}^2=289$ $\Rightarrow \text{AC}=17$ $\sin\text{A}=\frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{15}{17}$ $\sec\text{A}=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{17}{8}$
In $\triangle\text{ABC},$ $\text{AC}^2=\text{AB}^2+\text{BC}^2$ $\Rightarrow\text{AC}^2=(8)^2+(15)^2$ $\Rightarrow\text{AC}^2=64+225$ $\Rightarrow\text{AC}^2=289$ $\Rightarrow \text{AC}=17$ $\sin\text{A}=\frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{15}{17}$ $\sec\text{A}=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{17}{8}$
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