Question

Given $15\cot A=8$, find $\sin A$ and $\sec A$.

Answer 1

Let $ABC$ be a right angled $△$ right angled at $B$.

$\cot A=\frac{8}{15}$

Let $AB=8x$ and $BC$ will be $15x$.

$A{C}^2=A{B}^2+B{C}^2$

$\Rightarrow A{C}^2=64x^2+225x^2$

$\Rightarrow AC=\sqrt{289x^2}=17x$

$\Rightarrow \sin A=\frac{BC}{AC}=\frac{15x}{17x}=\frac{15}{17}$

$\Rightarrow \sec A=\frac{AC}{AB}=\frac{17x}{8x}=\frac{17}{8}$